Phương pháp giải bài toán tính tích phân hàm phân thức hữu tỉ

A. DẠNG : \(I=\int\limits_\alpha ^\beta \frac{{P(x)}}{{{\rm{ax + b}}}}dx\) \( \left( {a \ne 0} \right) \)

* Chú ý đến công thức : \(\int\limits_\alpha ^\beta {\frac{m}{{{\rm{ax + b}}}}dx} = \frac{m}{a}\ln \left| {{\rm{ax + b}}} \right|\left| {\begin{array}{*{20}{c}}\beta \\\alpha \end{array}} \right.\).
Và nếu bậc của P(x) cao hơn hoắc bằng 2 thì ta chia tử cho mẫu dẫn đến \(\int\limits_\alpha ^\beta {\frac{{P(x)}}{{{\rm{ax + b}}}}dx = \int\limits_\alpha ^\beta {Q(x) + \frac{m}{{{\rm{ax + b}}}}dx = \int\limits_\alpha ^\beta {Q(x)dx} + m\int\limits_\alpha ^\beta {\frac{1}{{{\rm{ax + b}}}}dx} \quad } \quad } \)

Ví dụ 1 : Tính tích phân : I= \(\int\limits_1^2 {\frac{{{x^3}}}{{2x + 3}}dx} \)

Giải

Ta có : \(f(x) = \frac{{{x^3}}}{{2x + 3}} = \frac{1}{2}{x^2} - \frac{3}{4}x + \frac{9}{8} - \frac{{27}}{8}\frac{1}{{2x + 3}}\)
Do đó : \(\int\limits_1^2 {\frac{{{x^3}}}{{2x + 3}}dx} = \int\limits_1^2 {\left( {\frac{1}{2}{x^2} - \frac{3}{4}x + \frac{9}{8} - \frac{{27}}{8}\frac{1}{{2x + 3}}} \right)dx} = \left( {\frac{1}{3}{x^3} - \frac{3}{8}{x^2} + \frac{9}{8}x - \frac{{27}}{{16}}\ln \left| {2x + 3} \right|} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = } \right. - \frac{{13}}{6} - \frac{{27}}{{16}}\ln 35\)

Ví dụ 2: Tính tích phân : I= \(\int\limits_{\sqrt 5 }^3 {\frac{{{x^2} - 5}}{{x + 1}}dx} \)

Giải

Ta có : f(x)=\(\frac{{{x^2} - 5}}{{x + 1}} = x - 1 - \frac{4}{{x + 1}}\).
Do đó : \(\int\limits_{\sqrt 5 }^3 {\frac{{{x^2} - 5}}{{x + 1}}dx} \) \( = \int\limits_{\sqrt 5 }^3 {\left( {x - 1 - \frac{4}{{x + 1}}} \right)dx} \) \(= \left( {\frac{1}{2}{x^2} - x - 4\ln \left| {x + 1} \right|} \right)\left| {\begin{array}{*{20}{c}}3\\{\sqrt 5 }\end{array} = } \right.\sqrt 5 - 1 + 4\ln \left( {\frac{{\sqrt 5 + 1}}{4}} \right)\)

B. DẠNG : \(\int\limits_\alpha ^\beta {\frac{{P(x)}}{{{\rm{a}}{{\rm{x}}^{\rm{2}}} + bx + c}}dx} \)

1. Tam thức : \(f(x) = {\rm{a}}{{\rm{x}}^{\rm{2}}} + bx + c\) có hai nghiệm phân biệt
Công thức cần lưu ý : \(\int\limits_\alpha ^\beta {\frac{{u'(x)}}{{u(x)}}dx} = \ln \left| {u(x)} \right|\left| {\begin{array}{*{20}{c}}\beta \\\alpha \end{array}} \right.\)
Ta có hai cách:
Cách 1: ( Hệ số bất định )
Cách 2: ( Nhẩy tầng lầu )

Ví dụ 3: Tính tích phân : I= \(\int\limits_0^1 {\frac{{4x + 11}}{{{x^2} + 5x + 6}}dx} \).

Giải

Cách 1: ( Hệ số bất định )
Ta có : f(x)=\(\frac{{4x + 11}}{{{x^2} + 5x + 6}} = \frac{{4x + 11}}{{(x + 2)(x + 3)}} \) \(= \frac{A}{{x + 2}} + \frac{B}{{x + 3}} = \frac{{A\left( {x + 3} \right) + B\left( {x + 2} \right)}}{{(x + 2)(x + 3)}}\)
Thay \(x=-2\) vào hai tử số : \(3=A\) và thay \(x=-3\) vào hai tử số : \(-1= -B\) suy ra \(B=1\) Do đó : \( f(x)=\frac{3}{{x + 2}} + \frac{1}{{x + 3}}\)
Vậy : \(\int\limits_0^1 {\frac{{4x + 11}}{{{x^2} + 5x + 6}}dx} \) \(= \int\limits_0^1 {\left( {\frac{3}{{x + 2}} + \frac{1}{{x + 3}}} \right)dx} \) \(= \left( {3\ln \left| {x + 2} \right| + \ln \left| {x + 3} \right|} \right)\left| {\begin{array}{*{20}{c}}1\\0\end{array} = 2\ln 3 - \ln 2} \right.\)
Cách 2: ( Nhẩy tầng lầu )
Ta có : f(x)=\(\frac{{2\left( {2x + 5} \right) + 1}}{{{x^2} + 5x + 6}} = 2.\frac{{2x + 5}}{{{x^2} + 5x + 6}} + \frac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\) \( = 2.\frac{{2x + 5}}{{{x^2} + 5x + 6}} + \frac{1}{{x + 2}} - \frac{1}{{x + 3}}\)
Do đó : \(I=\int\limits_0^1 {f(x)dx} = \int\limits_0^1 {\left( {2.\frac{{2x + 5}}{{{x^2} + 5x + 6}} + \frac{1}{{x + 2}} - \frac{1}{{x + 3}}} \right)dx = \left( {2\ln \left| {{x^2} + 5x + 6} \right| + \ln \left| {\frac{{x + 2}}{{x + 3}}} \right|} \right)\left| {\begin{array}{*{20}{c}}1\\0\end{array} = 2\ln 3 - \ln 2} \right.} \)

2. Tam thức : \(f(x) = {\rm{a}}{{\rm{x}}^{\rm{2}}} + bx + c\) có hai nghiệm kép
Công thức cần chú ý : \(\int\limits_\alpha ^\beta {\frac{{u'(x)dx}}{{u(x)}} = \ln \left( {u(x)} \right)\left| {\begin{array}{*{20}{c}}\beta \\\alpha \end{array}} \right.} \)
Thông thừơng ta đặt \((x+b/2a)=t\) .
Ví dụ 4 : Tính tích phân sau : I= \(\int\limits_0^3 {\frac{{{x^3}}}{{{x^2} + 2x + 1}}dx} \)

Giải

Ta có : \(\int\limits_0^3 {\frac{{{x^3}}}{{{x^2} + 2x + 1}}dx} = \int\limits_0^3 {\frac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx} \)
Đặt : \( t=x+1\) suy ra : \(dx=dt\) ; \(x=t-1\) và:
khi \(x=0\) thì \(t=1\); khi \(x=3\) thì \(t=4\) .
Do đó : \(\int\limits_0^3 {\frac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx} = \int\limits_1^4 {\frac{{{{\left( {t - 1} \right)}^3}}}{{{t^2}}}dt} \) \(= \int\limits_1^4 {\left( {t - 3 + \frac{3}{t} - \frac{1}{{{t^2}}}} \right)dt} = \left( {\frac{1}{2}{t^2} - 3t + \ln \left| t \right| + \frac{1}{t}} \right)\left| {\begin{array}{*{20}{c}}4\\1\end{array} = } \right.2\ln 2 - \frac{3}{2}\)

Ví dụ 5: Tính tích phân sau : I= \(\int\limits_0^1 {\frac{{4x}}{{4{x^2} - 4x + 1}}dx} \)

Giải

Ta có : \(\frac{{4x}}{{4{x^2} - 4x + 1}} = \frac{{4x}}{{{{\left( {2x - 1} \right)}^2}}}\)
Đặt : \(t= 2x-1\) suy ra : \(dt = 2dx \to dx = \frac{1}{2}dt; \)
Ta có: \(\left\{ \begin{array}{l}x = 0 \leftrightarrow t = - 1\\x = 1 \leftrightarrow t = 1\end{array} \right.\)
Do đó : \(\int\limits_0^1 {\frac{{4x}}{{4{x^2} - 4x + 1}}dx} \) \(= \int\limits_0^1 {\frac{{4x}}{{{{\left( {2x - 1} \right)}^2}}}dx} = \int\limits_{ - 1}^1 {\frac{{4.\frac{1}{2}\left( {t + 1} \right)}}{{{t^2}}}\frac{1}{2}dt} = \int\limits_{ - 1}^1 {\left( {\frac{1}{t} + \frac{1}{{{t^2}}}} \right)dt} = \left( {\ln \left| t \right| - \frac{1}{t}} \right)\left| {\begin{array}{*{20}{c}}1\\{ - 1}\end{array} = - 2} \right.\)

3. Tam thức : \(f(x) = {\rm{a}}{{\rm{x}}^{\rm{2}}} + bx + c\) vô nghiệm :
Ta viết : f(x)= \(\frac{{P(x)}}{{a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} + {{\left( {\frac{{\sqrt { - \Delta } }}{{2a}}} \right)}^2}} \right]}} = \frac{{P(x)}}{{a\left( {{u^2} + {k^2}} \right)}};\left\{ \begin{array}{l}u = x + \frac{b}{{2a}}\\k = \frac{{\sqrt { - \Delta } }}{{2a}}\end{array} \right.\)
Khi đó : Đặt \(u= k\tan t\)

Ví dụ 6: Tính tích phân : I= \(\int\limits_0^2 {\frac{x}{{{x^2} + 4x + 5}}dx} \)

Giải

• Ta có : \(\int\limits_0^2 {\frac{x}{{{x^2} + 4x + 5}}dx} = \int\limits_0^2 {\frac{x}{{{{\left( {x + 2} \right)}^2} + 1}}dx} \)
• Đặt : x+2=tant , suy ra : dx=\(\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt;\; \Rightarrow \left\{ \begin{array}{l}x = 0 \leftrightarrow \tan t = 2\\x = 2 \leftrightarrow \tan t = 4\end{array} \right.\)
• Do đó : \(\int\limits_0^2 {\frac{x}{{{{\left( {x + 2} \right)}^2} + 1}}dx} = \int\limits_{{t_1}}^{{t_2}} {\frac{{\tan t - 2}}{{1 + {{\tan }^2}t}}\frac{{dt}}{{c{\rm{o}}{{\rm{s}}^2}t}}} = \int\limits_{{t_1}}^{{t_2}} {\left( {\frac{{\sin t}}{{c{\rm{ost}}}} - 2} \right)dt} = \left( { - \ln \left| {c{\rm{ost}}} \right| - 2t} \right)\left| {\begin{array}{*{20}{c}}{{t_2}}\\{{t_1}}\end{array}} \right.\left( 1 \right)\)
Từ : \(\left[ \begin{array}{l}\tan t = 2 \leftrightarrow 1 + {\tan ^2}t = 5 \leftrightarrow c{\rm{o}}{{\rm{s}}^2}t = \frac{1}{5} \to c{\rm{os}}{{\rm{t}}_{\rm{1}}} = \frac{1}{{\sqrt 5 }}\\\tan t = 4 \leftrightarrow 1 + {\tan ^2}t = 17 \leftrightarrow c{\rm{o}}{{\rm{s}}^2}t = \frac{1}{{17}} \to c{\rm{os}}{{\rm{t}}_{\rm{2}}} = \frac{1}{{\sqrt {17} }}\end{array} \right.\)
• Vậy : \(\left( { - \ln \left| {c{\rm{ost}}} \right| - 2t} \right)\left| {\begin{array}{*{20}{c}}{{t_2}}\\{{t_1}}\end{array}} \right. = - \left[ {\left( {\ln \left| {c{\rm{os}}{{\rm{t}}_{\rm{2}}}} \right| - 2{t_2}} \right) - \left( {\ln \left| {\cos {t_1}} \right| - 2{t_1}} \right)} \right] = - \ln \left| {\frac{{c{\rm{os}}{{\rm{t}}_{\rm{2}}}}}{{{\rm{cos}}{{\rm{t}}_{\rm{1}}}}}} \right| + 2\left( {{t_2} - {t_1}} \right)\)
• \( \Leftrightarrow - \ln \left| {\frac{{c{\rm{os}}{{\rm{t}}_{\rm{2}}}}}{{{\rm{cos}}{{\rm{t}}_{\rm{1}}}}}} \right| + 2\left( {{t_2} - {t_1}} \right) = 2\left( {{\rm{arctan4 - arctan2}}} \right) - \ln \left| {\frac{1}{{\sqrt {17} }}.\sqrt 5 } \right| = 2\left( {{\rm{arctan4 - arctan2}}} \right) - \frac{1}{2}\ln \frac{5}{{17}}\)

Ví dụ 7: Tính tích phân sau : I= \(\int\limits_0^2 {\frac{{{x^3} + 2{x^2} + 4x + 9}}{{{x^2} + 4}}dx} \)

Giải

• Ta có : \(\frac{{{x^3} + 2{x^2} + 4x + 9}}{{{x^2} + 4}} = x + 2 + \frac{1}{{{x^2} + 4}}\)
• Do đó : \(\int\limits_0^2 {\frac{{{x^3} + 2{x^2} + 4x + 9}}{{{x^2} + 4}}dx} = \int\limits_0^2 {\left( {x + 2 + \frac{1}{{{x^2} + 4}}} \right)dx} \) \(= \left( {\frac{1}{2}{x^2} + 2x} \right)\left| {\begin{array}{*{20}{c}}2\\0\end{array} + \int\limits_0^2 {\frac{{dx}}{{{x^2} + 4}}} } \right. = 6 + J\) (1)
Tính tích phân J= \(\int\limits_0^2 {\frac{1}{{{x^2} + 4}}dx} \)
• Đặt : \(x=2\tan t\) suy ra :
\(dx=\frac{2}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt;\left\{ \begin{array}{l}x = 0 \to t = 0\\x = 2 \to t = \frac{\pi }{4}\end{array} \right. \leftrightarrow t \in \left[ {0;\frac{\pi }{4}} \right] \to c{\rm{ost > 0}}\)
• Khi đó : \(\int\limits_0^2 {\frac{1}{{{x^2} + 4}}dx} = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{1 + {{\tan }^2}t}}\frac{2}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {dt} = \frac{1}{2}t\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array} = \frac{\pi }{8}} \right.\)
• Thay vào (1) : \(I = 6 + \frac{\pi }{8}\)

C. DẠNG : \(\int\limits_\alpha ^\beta {\frac{{P(x)}}{{{\rm{a}}{{\rm{x}}^{\rm{3}}} + b{x^2} + cx + d}}dx} \)

1. Đa thức : f(x)=\({\rm{a}}{{\rm{x}}^{\rm{3}}} + b{x^2} + cx + d\;\left( {a \ne 0} \right)\) có một nghiệm bội ba
Công thức cần chú ý : \(\int\limits_\alpha ^\beta {\frac{1}{{{x^m}}}dx} = \frac{1}{{1 - m}}.\frac{1}{{{x^{m - 1}}}}\left| {\begin{array}{*{20}{c}}\beta \\\alpha \end{array}} \right.\)

Ví dụ 8: Tính tích phân : I= \(\int\limits_0^1 {\frac{x}{{{{\left( {x + 1} \right)}^3}}}dx} \)

Giải

Cách 1:
• Đặt : x+1=t , suy ra x=t-1 và : khi x=0 thì t=1 ; khi x=1 thì t=2
• Do đó : \(\int\limits_0^1 {\frac{x}{{{{\left( {x + 1} \right)}^3}}}dx} = \int\limits_1^2 {\frac{{t - 1}}{{{t^3}}}dt} = \int\limits_1^2 {\left( {\frac{1}{{{t^2}}} - \frac{1}{{{t^3}}}} \right)dt} = \left( { - \frac{1}{t} + \frac{1}{2}\frac{1}{{{t^2}}}} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{1}{8}} \right.\)
Cách 2:
• Ta có : \(\frac{x}{{{{\left( {x + 1} \right)}^3}}} = \frac{{\left( {x + 1} \right) - 1}}{{{{\left( {x + 1} \right)}^3}}} = \frac{1}{{{{\left( {x + 1} \right)}^2}}} - \frac{1}{{{{\left( {x + 1} \right)}^3}}}\)
• Do đó : \(\int\limits_0^1 {\frac{x}{{{{\left( {x + 1} \right)}^3}}}dx} = \int\limits_0^1 {\left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}} - \frac{1}{{{{\left( {x + 1} \right)}^3}}}} \right]} dx = \left[ { - \frac{1}{{x + 1}} + \frac{1}{2}\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]\left| {\begin{array}{*{20}{c}}1\\0\end{array} = \frac{1}{8}} \right.\)

Ví dụ 9: Tính tích phân : I=\(\int\limits_{ - 1}^0 {\frac{{{x^4}}}{{{{\left( {x - 1} \right)}^3}}}dx} \).

Giải

Đặt : \(x-1=t\) , suy ra : \(x=t+1\) và :
khi \(x=-1\) thì \(t=-2\) và khi \(x=0\) thì \(t=-1\).
Do đó : \(\int\limits_{ - 1}^0 {\frac{{{x^4}}}{{{{\left( {x - 1} \right)}^3}}}dx} = \int\limits_{ - 2}^{ - 1} {\frac{{{{\left( {t + 1} \right)}^4}}}{{{t^3}}}dt} = \int\limits_{ - 2}^{ - 1} {\frac{{{t^4} + 4{t^3} + 6{t^2} + 4t + 1}}{{{t^3}}}dt} = \int\limits_{ - 2}^{ - 1} {\left( {t + 4 + \frac{6}{t} + \frac{4}{{{t^2}}} + \frac{1}{{{t^3}}}} \right)dt} \)
\( \Leftrightarrow \int\limits_{ - 2}^{ - 1} {\left( {t + 4 + \frac{6}{t} + \frac{4}{{{t^2}}} + \frac{1}{{{t^3}}}} \right)dt} = \left( {\frac{1}{2}{t^2} + 4t + 6\ln \left| t \right| - \frac{4}{t} - \frac{1}{2}\frac{1}{{{t^2}}}} \right)\left| {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\end{array} = \frac{{33}}{8} - 6\ln 2} \right.\)

2. Đa thức : \(f(x)={\rm{a}}{{\rm{x}}^{\rm{3}}} + b{x^2} + cx + d\;\left( {a \ne 0} \right)\) có hai nghiệm :
Có hai cách giải : Hệ số bất định và phương pháp nhẩy tầng lầu

Ví dụ 10: Tính tích phân sau : I= \(\int\limits_2^3 {\frac{1}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^3}}}dx} \)

Giải

Cách 1. ( Phương pháp hệ số bất định )
• Ta có : \(\frac{1}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{\left( {x + 1} \right)}} + \frac{C}{{{{\left( {x + 1} \right)}^2}}} = \frac{{A{{\left( {x + 1} \right)}^2} + B\left( {x - 1} \right)\left( {x + 1} \right) + C\left( {x - 1} \right)}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}\)
• Thay hai nghiệm mẫu số vào hai tử số : \(\left\{ \begin{array}{l}1 = 4A\\1 = - 2C\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = \frac{1}{4}\\C = - \frac{1}{2}\end{array} \right.\).
Khi đó (1) \( \Leftrightarrow \frac{{\left( {A + B} \right){x^2} + \left( {2A + C} \right)x + A - B - C}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} \Rightarrow A - B - C = 1 \Leftrightarrow B = A - C - 1 = \frac{1}{4} + \frac{1}{2} - 1 = - \frac{1}{4}\)
• Do đó : \(\int\limits_2^3 {\frac{1}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}dx} = \int\limits_2^3 {\left( {\frac{1}{4}.\frac{1}{{x - 1}} + \frac{1}{4}.\frac{1}{{\left( {x + 1} \right)}} - \frac{1}{2}\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} \)
\( \Leftrightarrow I = \left[ {\frac{1}{4}\ln \left( {x - 1} \right)\left( {x + 1} \right) + \frac{1}{2}.\frac{1}{{\left( {x + 1} \right)}}} \right]\left| {\begin{array}{*{20}{c}}3\\2\end{array} = \frac{1}{4}\ln 8 = \frac{3}{4}\ln 2} \right.\)

Cách 2:
• Đặt : t=x+1, suy ra : x=t-1 và khi x=2 thì t=3 ; khi x=3 thì t=4 .
• Khi đó : I=\(\int\limits_2^3 {\frac{1}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}dx} = \int\limits_3^4 {\frac{{dt}}{{{t^2}\left( {t - 2} \right)}} = \frac{1}{2}\int\limits_3^4 {\frac{{t - \left( {t - 2} \right)}}{{{t^2}\left( {t - 2} \right)}}dt} = \frac{1}{2}\left( {\int\limits_2^4 {\frac{1}{{t\left( {t - 2} \right)}}dt - \int\limits_3^4 {\frac{1}{t}dt} } } \right)} \)
\( \Leftrightarrow I = \frac{1}{2}\left( {\frac{1}{2}\int\limits_2^4 {\left( {\frac{1}{{t - 2}} - \frac{1}{t}} \right)dt - \int\limits_3^4 {\frac{1}{t}dt} } } \right) = \left( {\frac{1}{4}\ln \left| {\frac{{t - 2}}{t}} \right| - \frac{1}{2}\ln \left| t \right|} \right)\left| {\begin{array}{*{20}{c}}4\\3\end{array} = } \right.\frac{3}{4}\ln 2\)
Hoặc :
\(\frac{1}{{{t^3} - 2{t^2}}} = \frac{{\left( {3{t^2} - 4t} \right)}}{{{t^3} - 2{t^2}}} - \frac{1}{4}\left( {\frac{{3{t^2} - 4t - 4}}{{{t^3} - 2{t^2}}}} \right) = \left[ {\frac{{3{t^2} - 4t}}{{{t^3} - 2{t^2}}} - \frac{1}{4}\frac{{\left( {3t + 2} \right)}}{{{t^2}}}} \right] = \frac{{3{t^2} - 4t}}{{{t^3} - 2{t^2}}} - \frac{1}{4}\left( {\frac{3}{t} + \frac{2}{{{t^2}}}} \right)\)
• Do đó : I=\(\int\limits_3^4 {\left( {\frac{{3{t^2} - 4t}}{{{t^3} - 2{t^2}}} - \frac{1}{4}\left( {\frac{3}{t} + \frac{2}{{{t^2}}}} \right)} \right)dt = \left( {\ln \left| {{t^3} - 2{t^2}} \right| - \frac{1}{4}\left( {3\ln \left| t \right| - \frac{2}{t}} \right)} \right)\left| {\begin{array}{*{20}{c}}4\\3\end{array} = } \right.} \frac{3}{4}\ln 2\)
Hoặc :
\(\frac{1}{{{t^2}\left( {t - 2} \right)}} = \frac{1}{4}\left( {\frac{{{t^2} - \left( {{t^2} - 4} \right)}}{{{t^2}\left( {t - 2} \right)}}} \right) = \frac{1}{4}\left( {\frac{1}{{t - 2}} - \frac{{t + 2}}{{{t^2}}}} \right) = \frac{1}{4}\left( {\frac{1}{{t - 2}} - \frac{1}{t} - \frac{2}{{{t^2}}}} \right)\)
• Do đó : I=\(\frac{1}{4}\int\limits_3^4 {\left( {\frac{1}{{t - 2}} - \frac{1}{t} - \frac{2}{{{t^2}}}} \right)dt = \frac{1}{4}\left( {\ln \left| {\frac{{t - 2}}{t}} \right| + \frac{2}{t}} \right)\left| {\begin{array}{*{20}{c}}4\\3\end{array}} \right.} = \frac{1}{4}\left( {\ln \frac{1}{2} + \frac{1}{2} - \ln \frac{1}{3} - \frac{2}{3}} \right) = \frac{1}{4}\left( {\ln 3 - \ln 2 - \frac{1}{6}} \right)\)

Ví dụ 11: Tính tích phân sau : I= \(\int\limits_2^3 {\frac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}dx} \)

Giải

Đặt : \(x-1=t\), suy ra: \(x=t+1\), \(dx=dt\) và: khi \(x=2\) thì \(t=1\); \(x=3\) thì \(t=2\).
Do đó : \(\int\limits_2^3 {\frac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}dx} = \int\limits_1^2 {\frac{{{{\left( {t + 1} \right)}^2}}}{{{t^2}\left( {t + 3} \right)}}dt} = \int\limits_1^2 {\frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}}dt} \)
Cách 1; ( Hệ số bất định )
Ta có :\(\frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}} = \frac{{At + B}}{{{t^2}}} + \frac{C}{{t + 3}} = \frac{{\left( {At + B} \right)\left( {t + 3} \right) + C{t^2}}}{{{t^2}\left( {t + 3} \right)}} = \frac{{\left( {A + C} \right){t^2} + \left( {3A + B} \right)t + 3B}}{{{t^2}\left( {t + 3} \right)}}\)
Đồng nhất hệ số hai tử số : \(\left\{ \begin{array}{l}A + C = 1\\3A + B = 2\\3B = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}B = \frac{1}{3}\\A = \frac{5}{9}\\C = \frac{4}{9}\end{array} \right. \Rightarrow \frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}} = \frac{1}{9}\frac{{t + 3}}{{{t^2}}} + \frac{4}{9}\frac{1}{{t + 3}}\)
Do đó : \(\int\limits_1^2 {\frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}}dt} = \int\limits_1^2 {\left( {\frac{1}{9}\left( {\frac{1}{t} + \frac{3}{{{t^2}}}} \right) + \frac{4}{9}\left( {\frac{1}{{t + 3}}} \right)} \right)dt} = \left( {\frac{1}{9}\left( {\ln \left| t \right| - \frac{3}{t}} \right) + \frac{4}{9}\ln \left| {t + 3} \right|} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = } \right.\frac{{17}}{6} + \frac{4}{9}\ln 5 - \frac{7}{9}\ln 2\)

Cách 2:
• Ta có : \(\frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}} = \frac{1}{3}\left( {\frac{{3{t^2} + 6t + 3}}{{{t^3} + 3{t^2}}}} \right) = \frac{1}{3}\left[ {\frac{{3{t^2} + 6t}}{{{t^3} + 3{t^2}}} + \frac{3}{{{t^2}\left( {t + 3} \right)}}} \right] = \frac{1}{3}\left[ {\left( {\frac{{3{t^2} + 6t}}{{{t^3} + 3{t^2}}}} \right) + \frac{1}{9}\left( {\frac{{{t^2} - \left( {{t^2} - 9} \right)}}{{{t^2}\left( {t + 3} \right)}}} \right)} \right]\)\( = \frac{1}{3}\left( {\frac{{3{t^2} + 6t}}{{{t^3} + 3{t^2}}}} \right) + \frac{1}{9}\frac{1}{{t + 3}} - \frac{1}{9}\frac{{t - 3}}{{{t^2}}} = \frac{1}{3}\left[ {\left( {\frac{{3{t^2} + 6t}}{{{t^3} + 3{t^2}}}} \right) + \frac{1}{9}\frac{1}{{t + 3}} - \frac{1}{9}\left( {\frac{1}{t} - \frac{3}{{{t^2}}}} \right)} \right]\)
• Vậy : \(\int\limits_1^2 {\frac{{{t^2} + 2t + 1}}{{{t^2}\left( {t + 3} \right)}}dt} = \int\limits_1^2 {\left( {\frac{1}{3}\left( {\frac{{3{t^2} + 6t}}{{{t^3} + 3{t^2}}}} \right) + \frac{1}{9}\left( {\frac{1}{{t + 3}} - \frac{1}{t} + \frac{3}{{{t^2}}}} \right)} \right)dt} = \left[ {\frac{1}{3}\ln \left| {{t^3} + 3{t^2}} \right| + \frac{1}{{27}}\left( {\ln \left| {\frac{{t + 3}}{t}} \right| - \frac{3}{t}} \right)} \right]\left| {\begin{array}{*{20}{c}}2\\1\end{array}} \right.\)
• Do đó I= \(\frac{{17}}{6} + \frac{4}{9}\ln 5 - \frac{7}{9}\ln 2\)
3. Đa thức : f(x)=\({\rm{a}}{{\rm{x}}^{\rm{3}}} + b{x^2} + cx + d\;\left( {a \ne 0} \right)\) có ba nghiệm :

Ví dụ 12: Tính tích phân sau : I= \(\int\limits_2^3 {\frac{1}{{x\left( {{x^2} - 1} \right)}}dx} \)

Giải

Cách 1: ( Hệ số bất định )
• Ta có : f(x)=\(\frac{1}{{x\left( {{x^2} - 1} \right)}} = \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} = \frac{{A\left( {{x^2} - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\)
• Đồng nhất hệ số hai tử số bằng cách thay các nghiệm : \(x=0\); \(x=1\) và \(x=-1\) vào hai tử ta có : \(\left\{ \begin{array}{l}x = 0 \to 1 = - A\\x = - 1 \to 1 = 2C\\x = 1 \to 1 = 2B\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = - 1\\B = \frac{1}{2}\\C = \frac{1}{2}\end{array} \right. \Rightarrow f(x) = - \frac{1}{x} + \frac{1}{2}\left( {\frac{1}{{x - 1}}} \right) + \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right)\)
• Vậy : \(\int\limits_2^3 {\frac{1}{{x\left( {{x^2} - 1} \right)}}dx} = \int\limits_2^3 {\left( {\frac{1}{2}\left( {\frac{1}{{x - 1}} + \frac{1}{{x + 1}}} \right) - \frac{1}{x}} \right)dx} = \left[ {\frac{1}{2}\left( {\ln \left( {x - 1} \right)\left( {x + 1} \right)} \right) - \ln \left| x \right|} \right]\left| {\begin{array}{*{20}{c}}3\\2\end{array} = \frac{5}{2}\ln 2 - \frac{3}{2}\ln 3} \right.\)

Cách 2: ( Phương pháp nhẩy lầu )
Ta có : \(\frac{1}{{x\left( {{x^2} - 1} \right)}} = \frac{{{x^2} - \left( {{x^2} - 1} \right)}}{{x\left( {{x^2} - 1} \right)}} = \frac{x}{{{x^2} - 1}} - \frac{1}{x} = \frac{1}{2}\frac{{2x}}{{{x^2} - 1}} - \frac{1}{x}\)
Do đó : \(\int\limits_2^3 {\frac{1}{{x\left( {{x^2} - 1} \right)}}dx} = \frac{1}{2}\int\limits_2^3 {\frac{{2xdx}}{{{x^2} - 1}} - \int\limits_2^3 {\frac{1}{x}dx} = \left( {\frac{1}{2}\ln \left( {{x^2} - 1} \right) - \ln x} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array} = } \right.} \frac{5}{2}\ln 2 - \frac{3}{2}\ln 3\)

Ví dụ 13: Tính tích phân sau : I=\(\int\limits_3^4 {\frac{{x + 1}}{{x\left( {{x^2} - 4} \right)}}dx} \)

Giải  

Cách 1:
Ta có : \(\frac{{x + 1}}{{x\left( {{x^2} - 4} \right)}} = \frac{{x + 1}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{A}{x} + \frac{B}{{x - 2}} + \frac{C}{{x + 2}} = \frac{{A\left( {{x^2} - 4} \right) + Bx\left( {x + 2} \right) + Cx\left( {x - 2} \right)}}{{x\left( {{x^2} - 4} \right)}}\)
Thay các nghiệm của mẫu số vào hai tử số :
• Khi x=0 : 1= -4A suy ra : A=-1/4
• Khi x=-2 : -1= 8C suy ra C=-1/8
• Khi x=2 : 3= 8B suy ra : B=3/8 .
Do đó : f(x) = \( - \frac{1}{4}\left( {\frac{1}{x}} \right) - \frac{1}{8}\left( {\frac{1}{{x - 2}}} \right) + \frac{3}{8}\left( {\frac{1}{{x + 2}}} \right)\)
Vậy : \(\int\limits_3^4 {\frac{{x + 1}}{{x\left( {{x^2} - 4} \right)}}dx} = - \frac{1}{4}\int\limits_2^3 {\frac{1}{x}dx - \frac{1}{8}\int\limits_2^3 {\frac{1}{{x - 2}}dx} + \frac{3}{8}\int\limits_2^3 {\frac{1}{{x + 2}}} dx = \left( { - \frac{1}{4}\ln \left| x \right| - \frac{1}{8}\ln \left| {x - 2} \right| + \frac{3}{8}\ln \left| {x + 2} \right|} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array} = } \right.} \)
\( = \frac{5}{8}\ln 3 - \frac{3}{8}\ln 5 - \frac{1}{4}\ln 2\)

Cách 2:
Ta có : \(\frac{{x + 1}}{{x\left( {{x^2} - 4} \right)}} = \frac{1}{{\left( {{x^2} - 4} \right)}} + \frac{1}{{x\left( {{x^2} - 4} \right)}} = \frac{1}{4}\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right) + \frac{1}{4}\left( {\frac{{{x^2} - \left( {{x^2} - 4} \right)}}{{x\left( {{x^2} - 4} \right)}}} \right) = \frac{1}{4}\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}} + \frac{1}{2}\frac{{2x}}{{{x^2} - 4}} - \frac{1}{x}} \right)\)
Do đó : \(\int\limits_3^4 {\frac{{x + 1}}{{x\left( {{x^2} - 4} \right)}}dx} = \frac{1}{4}\int\limits_3^4 {\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}} + \frac{1}{2}\frac{{2x}}{{{x^2} - 4}} - \frac{1}{x}} \right)dx = } \left[ {\frac{1}{4}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{1}{2}\ln \left( {{x^2} - 4} \right) - \ln \left| x \right|} \right]\left| {\begin{array}{*{20}{c}}4\\3\end{array}} \right.\)

Ví dụ 14: Tính tích phân sau : \(\int\limits_2^3 {\frac{{{x^2}}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}}dx} \)

Giải

Cách 1: ( Hệ số bất định )
\(\frac{{{x^2}}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}} = \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{x + 2}} = \frac{{A\left( {x + 1} \right)\left( {x + 2} \right) + B\left( {x - 1} \right)\left( {x + 2} \right) + C\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}}\)
Thay lần lượt các nghiệm mẫu số vào hai tử số :
• Thay : x=1 Ta cớ : 1=2A , suy ra : A=1/2
• Thay : x=-1 ,Ta có :1=-2B, suy ra : B=-1/2
• Thay x=-2 ,Ta có : 4= -5C, suy ra : C=-5/4
Do đó :
\(I=\int\limits_2^3 {\frac{{{x^2}}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}}dx} = \int\limits_2^3 {\left( {\frac{1}{2}\frac{1}{{x - 1}} - \frac{1}{2}\frac{1}{{x + 1}} - \frac{5}{4}\frac{1}{{x + 2}}} \right)} dx = \left[ {\frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{5}{4}\ln \left| {x + 2} \right|} \right]\left| {\begin{array}{*{20}{c}}3\\2\end{array}} \right. = \frac{1}{2}\ln \frac{3}{2}\)

Cách 2.( Nhẩy tầng lầu )
Ta có : \(\frac{{{x^2}}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}} = \frac{{{x^2} - 1 + 1}}{{\left( {{x^2} - 1} \right)\left( {x + 2} \right)}} = \frac{1}{{x + 2}} + \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{1}{{x + 2}} + \frac{1}{2}\frac{{x\left( {x + 1} \right) - \left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}}\)
\( = \frac{1}{{x + 2}} + \frac{1}{2}\left[ {\frac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}} - \frac{1}{{x + 1}}} \right] = \frac{1}{{x + 2}} + \frac{1}{2}\left[ {1 + \frac{1}{3}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 2}}} \right) - \frac{1}{{x + 1}}} \right]\)
Từ đó suy ra kết quả .

 

D. DẠNG \(\int\limits_\alpha ^\beta {\frac{{R\left( x \right)}}{{{\rm{a}}{{\rm{x}}^{\rm{4}}} + b{x^2} + c}}dx} \)

Những dạng này , gần đây trong các đề thi ít cho (Nhưng không hẳn là không cho ), nhưng tôi vẫn đưa ra đây một số đề thi đã thi trong những năm các trường ra đề thi riêng , mong các em học sinh khá , giỏi tham khảo để rút kinh nghiệm cho bản thân .
Sau đây là một số ví dụ minh họa

Ví dụ 1. Tính các tích phân sau :
a) \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 3x + 2} \right)}^2}}}dx} \)
b) \(\int\limits_{\frac{1}{2}}^1 {\frac{{1 + {x^2}}}{{1 + {x^3}}}dx} \)

Giải

a) \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 3x + 2} \right)}^2}}}dx} \) Ta có : \({x^2} + 3x + 2 = \left( {x + 1} \right)\left( {x + 2} \right) \Rightarrow f(x) = \frac{1}{{{{\left( {{x^2} + 3x + 2} \right)}^2}}} = \frac{1}{{{{\left[ {\left( {x + 1} \right)\left( {x + 2} \right)} \right]}^2}}} = {\left[ {\frac{1}{{\left( {x + 1} \right)}} - \frac{1}{{\left( {x + 2} \right)}}} \right]^2}\)
\( = \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 2} \right)}^2}}} - \frac{2}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 2} \right)}^2}}} - 2\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)\).
Vậy : \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 3x + 2} \right)}^2}}}dx} = \int\limits_0^1 {\left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 2} \right)}^2}}} - 2\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)} \right]dx = \left( { - \frac{1}{{x + 1}} - \frac{1}{{x + 2}} - 2\ln \left| {\frac{{x + 1}}{{x + 2}}} \right|} \right)} \begin{array}{*{20}{c}}1\\0\end{array} \) \(= \frac{2}{3} + 2\ln 3\)

 
b) \(\int\limits_{\frac{1}{2}}^1 {\frac{{1 + {x^2}}}{{1 + {x^3}}}dx} \)
Ta có : \(f(x) = \frac{{1 + {x^2}}}{{1 + {x^3}}} = \frac{{1 - x + {x^2} + x}}{{\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)}} = \frac{{1 - x + {x^2}}}{{\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)}} + \frac{x}{{\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)}}\)
\( \Leftrightarrow f(x) = \frac{1}{{1 + x}} + \frac{x}{{1 + {x^3}}} \Rightarrow \int\limits_{\frac{1}{2}}^1 {\left( {\frac{1}{{x + 1}} + \frac{1}{2}\frac{{2x}}{{1 + {x^3}}}} \right)} dx\)

Ví dụ 2. Tính các tích phân sau
a) \(\int\limits_1^{\sqrt 3 } {\frac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}}dx} \)
b) \(\int\limits_0^1 {\frac{{{x^4} + 1}}{{{x^6} + 1}}dx} \)

Giải

a) \(\int\limits_1^{\sqrt 3 } {\frac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}}dx} \) . Chia tử và mẫu cho \({x^2} \ne 0\), ta có :
\(f(x) = \frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}} - 1}} \Rightarrow \int\limits_1^{\sqrt 3 } {f(x)dx = \int\limits_1^{\sqrt 3 } {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)dx}}{{\left( {{x^2} + \frac{1}{{{x^2}}}} \right) - 1}}\quad \left( 1 \right)} } \)
Đặt : \(t = x + \frac{1}{x} \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {t^2} - 2,dt = \left( {1 - \frac{1}{{{x^2}}}} \right)dx \leftrightarrow \left[ \begin{array}{l}x = 1 \to t = 2\\x = \sqrt 3 \to t = \frac{4}{{\sqrt 3 }}\end{array} \right.\)
Vậy : \(\int\limits_1^{\sqrt 3 } {f(x)dx = \int\limits_2^{\frac{4}{{\sqrt 3 }}} {\frac{{dt}}{{{t^2} - 3}} = \int\limits_2^{\frac{4}{{\sqrt 3 }}} {\frac{1}{{\left( {t - \sqrt 3 } \right)\left( {t + \sqrt 3 } \right)}}dt = \frac{1}{{2\sqrt 3 }}\int\limits_2^{\frac{4}{{\sqrt 3 }}} {\left( {\frac{1}{{t - \sqrt 3 }} - \frac{1}{{t + \sqrt 3 }}} \right)dt} } } } \)
\(I = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{t - \sqrt 3 }}{{t + \sqrt 3 }}} \right|\left| {\begin{array}{*{20}{c}}{\frac{4}{{\sqrt 3 }}}\\2\end{array} = \frac{1}{{2\sqrt 3 }}\left( {\ln \frac{1}{7} - \ln \frac{{7 - 4\sqrt 3 }}{7}} \right) = \frac{1}{{2\sqrt 3 }}\ln \left( {7 + 4\sqrt 3 } \right)} \right.\))

b) \(\int\limits_0^1 {\frac{{{x^4} + 1}}{{{x^6} + 1}}dx} \).
Vì : \(\left[ \begin{array}{l}{x^6} - 1 = {\left( {{x^2}} \right)^3} - 1 = \left( {{x^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right)\\{x^6} - 1 = {\left( {{x^3}} \right)^2} - 1 = {t^2} - 1\left( {t = {x^3}} \right)\end{array} \right.\)
Cho nên : \(f(x) = \frac{{{x^4} + 1}}{{{x^6} + 1}} = \frac{{{x^4} - {x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}} - \frac{{{x^2}}}{{{{\left( {{x^3}} \right)}^2} + 1}} \Rightarrow \int\limits_0^1 {f(x)dx = \int\limits_0^1 {\left[ {\frac{1}{{{x^2} + 1}} - \frac{1}{3}\frac{{3{x^2}}}{{{{\left( {{x^3}} \right)}^2} + 1}}} \right]} } dx\)
Vậy : \(I = \arctan x\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. - \frac{1}{3}{\rm{arctan}}\left( {{\rm{3}}{{\rm{x}}^{\rm{2}}}} \right)\left| {\begin{array}{*{20}{c}}1\\0\end{array} = {\rm{arctan1 - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{arctan3}}} \right. = \frac{\pi }{4} - \frac{{\rm{1}}}{{\rm{3}}}{\rm{arctan3}}\)

Ví dụ 3. Tính các tích phân sau:
a) \(\int\limits_0^1 {\frac{{{x^2} + 1}}{{{x^4} + 1}}dx\quad \vee \int\limits_0^1 {\frac{{{x^2} - 1}}{{{x^4} + 1}}} } dx\)
b) \(\int\limits_1^2 {\frac{1}{{{x^4} + 1}}dx} \)

Giải

a. \(\int\limits_0^1 {\frac{{{x^2} + 1}}{{{x^4} + 1}}dx\quad \vee \int\limits_0^1 {\frac{{{x^2} - 1}}{{{x^4} + 1}}} } dx\).
Ta có : \(f(x) = \frac{{{x^2} + 1}}{{{x^4} + 1}} = \frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}},\quad g(x) = \frac{{{x^2} - 1}}{{{x^4} + 1}} = \frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}\).
Cho nên
Đặt : \(\left[ \begin{array}{l}t = x + \frac{1}{x} \Rightarrow dt = \left( {1 - \frac{1}{{{x^2}}}} \right)dx,{x^2} + \frac{1}{{{x^2}}} = {t^2} - 2,x = 1 \to t = 2,x = 2 \to t = \frac{5}{2}\\t = x - \frac{1}{x} \Rightarrow dt = \left( {1 + \frac{1}{{{x^2}}}} \right)dx,{x^2} + \frac{1}{{{x^2}}} = {t^2} + 2,x = 1 \to t = 0,x = 2 \to t = \frac{3}{2}\end{array} \right.\) .
Vậy: \( \int\limits_1^2 {f(x)dx = \int\limits_2^{\frac{5}{2}} {\left( {\frac{{dt}}{{{t^2} - 2}}} \right) = \int\limits_2^{\frac{5}{2}} {\frac{1}{{\left( {t - \sqrt 2 } \right)\left( {t + \sqrt 2 } \right)}}dt = \frac{1}{{2\sqrt 2 }}\int\limits_2^{\frac{5}{2}} {\left( {\frac{1}{{t - \sqrt 2 }} - \frac{1}{{t + \sqrt 2 }}} \right)dt = } } } } \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{t - \sqrt 2 }}{{t + \sqrt 2 }}} \right|\left| {\begin{array}{*{20}{c}}{\frac{5}{2}}\\2\end{array}} \right.\) \( \Leftrightarrow \int\limits_1^2 {g(x)dx = \int\limits_0^{\frac{3}{2}} {\frac{1}{{{t^2} + 2}}dt\quad \left( 1 \right)} } \) .
Đặt : \(t = \sqrt 2 \tan u \to dt = \sqrt 2 \frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}u}}du \leftrightarrow t = 0 \to u = 0,t = \frac{3}{2} \to u = {\rm{arctan}}\frac{{{\rm{3}}\sqrt 2 }}{4} = {u_1}\)
Do đó (1)\( \Leftrightarrow \int\limits_0^{{u_1}} {\frac{{\sqrt 2 du}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}u\left( {2 + 2{{\tan }^2}u} \right)}} = \int\limits_0^{{u_1}} {\frac{{\sqrt 2 }}{2}du = } } \frac{{\sqrt 2 }}{2}u\left| {\begin{array}{*{20}{c}}{{u_1}}\\0\end{array} = } \right.\frac{{\sqrt 2 }}{2}{u_1}\)

b) \(\int\limits_1^2 {\frac{1}{{{x^4} + 1}}dx} \) . Ta có : \(F(x) = \frac{1}{{{x^4} + 1}} = \frac{1}{2}\left( {\frac{{1 + {x^2} + 1 - {x^2}}}{{{x^4} + 1}}} \right) = \frac{1}{2}\left( {\frac{{{x^2} + 1}}{{{x^4} + 1}} - \frac{{{x^2} - 1}}{{{x^4} + 1}}} \right) = \frac{1}{2}\left( {f(x) - g(x)} \right)\)
Đã tính ở trên ( phần a)

Ví dụ 4. Tính các tích phân sau
a) \(\int\limits_1^2 {\frac{{{x^2} - 1}}{{\left( {{x^2} - 5x + 1} \right)\left( {{x^2} - 3x + 1} \right)}}dx} \)
b) \(\int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\frac{{dx}}{{{x^4} - 4{x^2} + 3}}} \)
c) \(\int\limits_1^{\frac{{1 - \sqrt 5 }}{2}} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}dx} \)
d) I =\(\int\limits_2^3 {\frac{{{x^7}}}{{1 + {x^8} - 2{x^4}}}} dx\)

Giải

a) \(\int\limits_1^2 {\frac{{{x^2} - 1}}{{\left( {{x^2} - 5x + 1} \right)\left( {{x^2} - 3x + 1} \right)}}dx} \).
Ta có : \(f(x) = \frac{{{x^2} - 1}}{{\left( {{x^2} - 5x + 1} \right)\left( {{x^2} - 3x + 1} \right)}} = \frac{{1 - \frac{1}{{{x^2}}}}}{{\left( {x + \frac{1}{x} - 5} \right)\left( {x + \frac{1}{{x - 3}}} \right)}} \Rightarrow \int\limits_1^2 {f(x)dx = \int\limits_1^2 {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)dx}}{{\left( {x + \frac{1}{x} - 5} \right)\left( {x + \frac{1}{x} - 3} \right)}}\quad \left( 1 \right)} } \)
Đặt : \(t = x + \frac{1}{x} \to dt = \left( {1 - \frac{1}{{{x^2}}}} \right)dx\;,x = 1 \to t = 2,x = 2 \to t = \frac{5}{2}\)
Vậy (1) trở thành : \(\int\limits_2^{\frac{5}{2}} {\frac{{dt}}{{\left( {t - 5} \right)\left( {t - 3} \right)}} = \frac{1}{2}\int\limits_2^{\frac{5}{2}} {\left( {\frac{1}{{t - 5}} - \frac{1}{{t - 3}}} \right)dt = \frac{1}{2}\ln \left| {\frac{{t - 5}}{{t - 3}}} \right|\left| {\begin{array}{*{20}{c}}{\frac{5}{2}}\\2\end{array}} \right. = \frac{1}{2}\left( {\ln 5 - \ln 3} \right) = \frac{1}{2}\ln \frac{5}{3}} } \)

b) \(\int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\frac{{dx}}{{{x^4} - 4{x^2} + 3}}} \).
Ta có : \(f(x) = \frac{1}{{{x^4} - 4{x^2} + 3}} = \frac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 3} \right)}} = \frac{1}{2}\left( {\frac{1}{{{x^2} - 3}} - \frac{1}{{{x^2} - 1}}} \right)\)
Do đó : \(\int\limits_{\frac{3}{2}}^{\frac{5}{2}} {f(x)dx = \int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\left( {\frac{1}{{{x^2} - 3}} - \frac{1}{{{x^2} - 1}}} \right)dx = I - J\quad \left( 1 \right)} } \)
Với : \(I = \int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\frac{1}{{{x^2} - 3}}dx = \int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\frac{1}{{\left( {x - \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)}}dx = \frac{1}{{2\sqrt 3 }}\int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\left( {\frac{1}{{x - \sqrt 3 }} - \frac{1}{{x + \sqrt 3 }}} \right)dx = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{x - \sqrt 3 }}{{x + \sqrt 3 }}} \right|\left| {\begin{array}{*{20}{c}}{\frac{5}{2}}\\{\frac{3}{2}}\end{array} = \frac{1}{{2\sqrt 3 }}\ln \frac{{37 - 20\sqrt 3 }}{{65\left( {7 - 4\sqrt 3 } \right)}}} \right.} } } \)
\(J = \int\limits_0^1 {\frac{1}{{{x^2} - 1}}dx = \int\limits_0^1 {\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}dx = \frac{1}{2}\int\limits_{\frac{3}{2}}^{\frac{5}{2}} {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right)dx = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right|\left| {\begin{array}{*{20}{c}}{\frac{5}{2}}\\{\frac{3}{2}}\end{array} = \frac{1}{2}\left( {\ln \frac{3}{7} - \ln \frac{1}{5}} \right)} \right.} } } = \frac{1}{2}\ln \frac{{15}}{7}\)

c) \(\int\limits_1^{\frac{{1 - \sqrt 5 }}{2}} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}dx} \).
Bạn hãy xem lại cách giải ví dụ 2-a . Chỉ khác là đặt : \(t = x - \frac{1}{x}\), sẽ ra kết quả .

d) I =\(\int\limits_2^3 {\frac{{{x^7}}}{{1 + {x^8} - 2{x^4}}}} dx = \int\limits_2^3 {\frac{{{x^4}}}{{{{\left( {{x^4} - 1} \right)}^2}}}{x^3}dx\quad \left( 1 \right)} \) Đặt : \(t = {x^4} - 1 \Rightarrow \left\{ \begin{array}{l}dt = 3{x^3}dx,x = 2 \to t = 15;x = 3 \to t = 80\\f(x)dx = \frac{1}{3}\frac{{{x^4}}}{{\left( {{x^4} - 1} \right)}}3{x^3}dx = \frac{1}{3}\frac{{\left( {t + 1} \right)}}{{{t^2}}}dt = \frac{1}{3}\left( {\frac{1}{t} + \frac{1}{{{t^2}}}} \right)dt\end{array} \right.\) Vậy : \(I = \int\limits_{15}^{80} {\frac{1}{3}\left( {\frac{1}{t} + \frac{1}{{{t^2}}}} \right)dt = \frac{1}{3}\left( {\ln \left| t \right| - \frac{1}{t}} \right)\left| {\begin{array}{*{20}{c}}{80}\\{15}\end{array} = \frac{1}{3}\ln \frac{{16}}{3} + \frac{{13}}{{720}}} \right.} \)

E. TRƯỜNG HỢP : \(\int\limits_\alpha ^\beta {\frac{{R\left( x \right)}}{{Q(x)}}dx} \quad \)( Với \(Q(x)\) có bậc cao hơn 4 )

Ở đây ta lưu ý : Đối với hàm phân thức hữu tỷ có bậc tử thấp hơn bậc mẫu tới hai bậc hoặc tinh ý nhận ra tính chất đặc biệt của hàm số dưới dấu tích phân mà có cách giải ngắn gọn hơn . Phương pháp chung là như vậy , nhưng chúng ta khéo léo hơn thì cách giải sẽ hay hơn .
Sau đây là minh họa bằng một số ví dụ:

Ví dụ 1. Tính các tích phân sau .
a.) \(\int\limits_1^2 {\frac{{dx}}{{x\left( {{x^4} + 1} \right)}}} \)
b) \(\int\limits_0^{\frac{1}{2}} {\frac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^2}\left( {x + 3} \right)}}dx\quad } \)

Giải

a) \(\int\limits_1^2 {\frac{{dx}}{{x\left( {{x^4} + 1} \right)}}} \) . Nếu theo cách phân tích bằng đồng nhất hệ số hai tử số thì ta có :
\(f(x) = \frac{1}{{x\left( {{x^4} + 1} \right)}} = \frac{A}{x} + \frac{{B{x^3} + C{x^2} + Dx + E}}{{{x^4} + 1}} = \frac{{A\left( {{x^4} + 1} \right) + x\left( {B{x^3} + C{x^2} + Dx + E} \right)}}{{x\left( {{x^4} + 1} \right)}} = \) \( \Leftrightarrow f(x) = \frac{{\left( {A + B} \right){x^4} + C{x^3} + D{x^2} + {\rm{Ex + A}}}}{{x\left( {{x^4} + 1} \right)}} \Rightarrow \left\{ \begin{array}{l}A + B = 0\\C = 0,D = 0\\E = 0\\A = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = 1\\B = - 1\\C = 0,D = 0,\\E = 0\end{array} \right. \Rightarrow f(x) = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}\)
Nhưng nếu ta tinh ý thì cách làm sau sẽ hay hơn .
Vì x và \({x^3}\) cách nhau 3 bậc , mặt khác \(x \in \left[ {1;2} \right] \Rightarrow x \ne 0\). Cho nên ta nhân tử và mẫu với \({x^3} \ne 0\).
Khi đó \(f(x) = \frac{{{x^3}}}{{{x^4}\left( {{x^4} + 1} \right)}}\).
Mặt khác \(d\left( {{x^4}} \right) = 4{x^3}dx \Leftrightarrow dt = 4{x^3}dx\quad \left( {t = {x^4}} \right)\), cho nên :
\(f(x)dx = \frac{1}{3}\frac{{3{x^3}dx}}{{{x^4}\left( {{x^4} + 1} \right)}} = \frac{1}{3}\frac{{dt}}{{t\left( {t + 1} \right)}} = \frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{t + 1}}} \right) = f(t)\).
Bài toán trở nên đơn giản hơn rất nhiều . ( Các em giải tiếp )

b) \(\int\limits_0^{\frac{1}{2}} {\frac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 3} \right)}}dx\quad } \)
Nhận xét :
* Nếu theo cách hướng dẫn chung ta làm như sau :
- \(f(x) = \frac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^3}\left( {x + 3} \right)}} = \frac{A}{{{{\left( {x - 1} \right)}^3}}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x - 1}} + \frac{D}{{x + 3}}\)
- Sau đó quy đồng mẫu số , đồng nhất hệ số hai tử số , ta có : \(A = \frac{1}{2},B = \frac{3}{8},C = - D = \frac{5}{{32}}\)
Do vậy : \(I = \int\limits_0^{\frac{1}{2}} {\left( {\frac{1}{{2{{\left( {x - 1} \right)}^3}}} + \frac{3}{{8{{\left( {x - 1} \right)}^2}}} + \frac{5}{{32\left( {x - 1} \right)}} - \frac{5}{{32\left( {x + 3} \right)}}} \right)dx} \) \( = \left[ { - \frac{1}{{8{{\left( {x - 1} \right)}^2}}} - \frac{3}{{8\left( {x - 1} \right)}} + \frac{5}{{32}}\ln \left| {x - 1} \right| - \frac{5}{{32}}\ln \left| {x + 3} \right|} \right]\left| {\begin{array}{*{20}{c}}{\frac{1}{2}}\\0\end{array} = } \right.\frac{5}{{32}}\ln \frac{1}{{28}}\)

Ví dụ 2. Tính các tích phân sau :
a. \(\int\limits_2^3 {\frac{{{x^4} - 1}}{{{x^6} - 1}}dx} \)
b. \(\int\limits_1^2 {\frac{{{x^2} + 1}}{{{x^6} + 1}}dx} \)
c. \(\int\limits_1^2 {\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}} \)
d. \(\int\limits_0^1 {\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx} \)
e. \(\int\limits_0^1 {\frac{{{x^4} + 3{x^2} + 1}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx} \)
f. \(\int\limits_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx} \)

Giải

a) \(\int\limits_1^2 {\frac{{{x^4} - 1}}{{{x^6} - 1}}dx} = \int\limits_1^2 {\left( {\frac{{{x^4} + {x^2} + 1}}{{\left( {{x^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right)}} - \frac{{{x^2} + 2}}{{\left[ {{{\left( {{x^3}} \right)}^2} - 1} \right]}}} \right)dx} = \int\limits_2^3 {\frac{1}{{{x^2} - 1}}dx + \int\limits_2^3 {\left( {\frac{{{x^2}}}{{\left[ {{{\left( {{x^3}} \right)}^2} - 1} \right]}} + \frac{1}{{{x^3} - 1}} - \frac{1}{{{x^3} + 1}}} \right)} } dx\)
Tính J : J= artanx\(\left| {\begin{array}{*{20}{c}}3\\2\end{array}} \right. = {\rm{artan3 - artan2}}\).
Tính K . Đặt \(t = {x^3} \Rightarrow \left\{ \begin{array}{l}dt = 3{x^2}dx,x = 2 \to t = 8;x = 3 \to t = 27\\g(x)dx = \frac{{{x^2}}}{{{x^3} - 1}}dx = \frac{1}{3}\frac{{dt}}{{\left( {{t^2} - 1} \right)}} = \frac{1}{3}\frac{1}{2}\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)dt\end{array} \right.\)
Do đó : K=\(\int\limits_2^3 {g(x)dx} = \frac{1}{6}\int\limits_8^{27} {\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)dt} = \frac{1}{6}\left( {\ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} \right)\left| {\begin{array}{*{20}{c}}{27}\\8\end{array} = \frac{1}{6}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|\left| {\begin{array}{*{20}{c}}{27}\\8\end{array} = } \right.} \right.\frac{1}{6}\ln \frac{{117}}{{98}}\)
Tính E=\(\int\limits_2^3 {\frac{1}{{{x^3} - 1}}dx} = \int\limits_2^3 {\frac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx} \) Ta có : \(h(x) = \frac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2} - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{{x^2} - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\)
\( = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{1}{2}\left( {\frac{{2x + 1}}{{{x^2} + x + 1}} + \frac{1}{{{x^2} + x + 1}}} \right)\)
Vậy : \(I = \frac{1}{3}\int\limits_2^3 {\frac{{3{x^2}}}{{{x^3} - 1}}dx} - \frac{1}{2}\int\limits_2^3 {\frac{{\left( {2x + 1} \right)}}{{{x^2} + x + 1}}dx - \int\limits_2^3 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx} } \)
\( = \frac{1}{3}\ln \left( {{x^3} - 1} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array} - \frac{1}{2}\ln \left( {{x^2} + x + 1} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array} - F = \frac{1}{3}\ln \frac{{28}}{9} - \frac{1}{2}\ln \frac{{13}}{6} - F} \right.} \right.\quad \left( 2 \right)\)
Tính F : Đặt : \(x + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t \Rightarrow \left\{ \begin{array}{l}dx = \frac{{\sqrt 3 }}{2}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt\\x = 2 \to \tan t = \frac{5}{{\sqrt 3 }} \to t = a;x = 3 \to \tan t = \frac{{10}}{{\sqrt 3 }} \to t = b\end{array} \right.\)
Do đó F=\(\int\limits_a^b {\frac{{\frac{{\sqrt 3 }}{2}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt}}{{\frac{{\sqrt 3 }}{2}\left( {1 + {{\tan }^2}t} \right)}} = \int\limits_a^b {dt} = t\left| {\begin{array}{*{20}{c}}b\\a\end{array} = b - a\quad \left( {{\mathop{\rm t}\nolimits} {\rm{ant = }}\frac{{\rm{5}}}{{\sqrt {\rm{3}} }} \to t = a = {\rm{artan}}\frac{{\rm{5}}}{{\sqrt {\rm{3}} }};b = {\rm{artan}}\frac{{{\rm{10}}}}{{\sqrt {\rm{3}} }}} \right)} \right.} \)
Thay vào (2) ta có kết quả .

b) \(\int\limits_1^2 {\frac{{{x^2} + 1}}{{{x^6} + 1}}dx} = \int\limits_0^1 {\frac{{{x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}dx = \int\limits_1^2 {\frac{1}{{{{\left( {{x^2} - 1} \right)}^2} - {x^2}}}dx} } = \int\limits_1^2 {\frac{1}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}}dx} \)
Ta có : \(\frac{1}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{{\rm{Ax + B}}}}{{{x^2} + x + 1}} + \frac{{Cx + D}}{{{x^2} - x + 1}}\) \( = \frac{{\left( {A + C} \right){x^3} + \left( {B - A + C + D} \right){x^2} + \left( {A - B + C + D} \right)x + \left( {B + D} \right)}}{{{x^4} - {x^2} + 1}}\)
Đồng nhất hệ số hai tử số ta có hệ : \(\left\{ \begin{array}{l}A + C = 0\\B - A + C + D = 0\\A - B + C + D = 0\\B + D = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = - C\\1 - 2C = 0\\ - B + D = 0\\B + D = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = - \frac{1}{2}\\C = \frac{1}{2}\\D = \frac{1}{2}\\B = \frac{1}{2}\end{array} \right.\)
Vậy : \(I = \frac{1}{2}\left( {\int\limits_1^2 {\frac{{1 - x}}{{{x^2} + x + 1}}dx + \int\limits_1^2 {\frac{{x + 1}}{{{x^2} - x + 1}}dx} } } \right) = \frac{1}{2}\left( {J + K} \right)\left( 1 \right)\)
Tính \(J= \int\limits_1^2 {\frac{{ - x + 1}}{{{x^2} + x + 1}}dx} = - \frac{1}{2}\int\limits_1^2 {\frac{{2x + 1 - 3}}{{{x^2} + x + 1}}dx} = - \frac{1}{2}\int\limits_1^2 {\frac{{2x + 1}}{{{x^2} + x + 1}}dx} + \frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx = - \frac{1}{2}\ln \left| {{x^2} + x + 1} \right|\left| {\begin{array}{*{20}{c}}2\\1\end{array} + E\quad \left( 2 \right)} \right.} \) Tính \(E=\frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx} \),
Em hãy tự tính bằng cách đặt : \(x + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t\)
Tính \(K = \int\limits_1^2 {\frac{{x + 1}}{{{x^2} - x + 1}}dx} = \frac{1}{2}\int\limits_1^2 {\frac{{2x - 1 + 3}}{{{x^2} - x + 1}}dx} = \frac{1}{2}\int\limits_1^2 {\frac{{2x - 1}}{{{x^2} - x + 1}}dx} + \frac{3}{2}\int\limits_0^1 {\frac{1}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx = \frac{1}{2}\ln \left| {{x^2} - x + 1} \right|\left| {\begin{array}{*{20}{c}}2\\1\end{array} + F\quad \left( 2 \right)} \right.} \) Tính \(F=\frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx} \),
Em hãy tự tính bằng cách đặt : \(x - \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t\)

c) \(\int\limits_1^2 {\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}} = \frac{1}{3}\int\limits_1^2 {\frac{{3{x^3}}}{{{x^4}\left( {1 + {x^4}} \right)}}dx} = \frac{1}{3}\int\limits_1^2 {\left( {\frac{{d\left( {{x^4}} \right)}}{{{x^4}}} - \frac{{d\left( {{x^4}} \right)}}{{1 + {x^4}}}} \right) = \frac{1}{3}\ln \left( {\frac{{{x^4}}}{{1 + {x^4}}}} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{1}{3}\ln \frac{{32}}{{17}}} \right.} \)

d) \(\int\limits_0^1 {\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx} = \frac{1}{2}\int\limits_0^1 {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}2xdx} \quad \left( 1 \right)\).
Đặt : \(t = 1 + {x^2} \Rightarrow \left\{ \begin{array}{l}{x^2} = t - 1;dt = 2xdx\\x = 0 \to t = 1,x = 1 \to t = 2\end{array} \right.\)
Do đó \(I = \int\limits_1^2 {\frac{{t - 1}}{{{t^3}}}dt} = \int\limits_1^2 {\left( {\frac{1}{{{t^2}}} - \frac{1}{{{t^3}}}} \right)dt} = \left( { - \frac{1}{t} + \frac{1}{{4{t^2}}}} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{{13}}{{16}}} \right.\)

e) \(\int\limits_0^1 {\frac{{{x^4} + 3{x^2} + 1}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx} = \int\limits_0^1 {\left( {\frac{{{{\left( {1 + {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}} + \frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}} \right)dx} = \int\limits_0^1 {\frac{1}{{1 + {x^2}}}dx} + \int\limits_0^1 {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx} = J + K\left( 1 \right)\)
Tính \(J\): Bằng cách đặt \(x = \tan t \Rightarrow J = \frac{\pi }{4}\)
Tính \(K= \int\limits_0^1 {\left( {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} - \frac{1}{{{{\left( {1 + {x^2}} \right)}^3}}}} \right)dx} = E + F\left( 2 \right)\)
Tính \(E\): Bằng cách đặt \(\begin{array}{l}x = \tan t \leftrightarrow \left\{ \begin{array}{l}dx = \frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt\\x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}\end{array} \right.\\\end{array}\)
Vậy : \(E = \frac{1}{2}\int\limits_0^1 {{{\left( {\frac{1}{{1 + {x^2}}}} \right)}^2}dx = \frac{1}{2}} \int\limits_0^{\frac{\pi }{4}} {{{\left( {\frac{1}{{1 + {{\tan }^2}t}}} \right)}^2}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt = } \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{4}}}t}}}}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {c{\rm{o}}{{\rm{s}}^{\rm{2}}}tdt} \) \( = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + c{\rm{os2t}}} \right)dt} = \frac{1}{4}\left( {t + \frac{1}{2}\sin 2t} \right)\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array} = \frac{1}{4}\left( {\frac{\pi }{4} + \frac{1}{2}} \right) = \frac{{\pi + 2}}{{16}}} \right.\)
Tính \(F\). Tương tự như tính \(E\); Bằng cách đặt \(\begin{array}{l}x = \tan t \leftrightarrow \left\{ \begin{array}{l}dx = \frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt\\x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}\end{array} \right.\\\end{array}\)
Vậy : \(F = \frac{1}{2}\int\limits_0^1 {{{\left( {\frac{1}{{1 + {x^2}}}} \right)}^3}dx = \frac{1}{2}} \int\limits_0^{\frac{\pi }{4}} {{{\left( {\frac{1}{{1 + {{\tan }^2}t}}} \right)}^3}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt = } \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{6}}}t}}}}\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {c{\rm{o}}{{\rm{s}}^{\rm{4}}}tdt} \)
\( = \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {{{\left( {1 + c{\rm{os2t}}} \right)}^2}dt} = \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + 2c{\rm{os}}2t + \frac{{1 + c{\rm{os4t}}}}{2}} \right)} dt\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array} = } \right.\) \(\frac{1}{{16}}\int\limits_0^{\frac{\pi }{4}} {\left( {3 + 4\cos 2t + c{\rm{os4t}}} \right)dt = } \frac{1}{{16}}\left( {3t + 2\sin 2t + \frac{1}{4}\sin 4t} \right)\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array} = \frac{1}{{16}}\left( {3\frac{\pi }{4} + 2} \right) = \frac{{3\pi + 8}}{{64}}} \right.\)

f) \(\int\limits_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx} = \int\limits_{\frac{1}{3}}^1 {{{\left( {\frac{{x - {x^3}}}{{{x^3}}}} \right)}^{\frac{1}{3}}}\frac{1}{{{x^3}}}dx} = \int\limits_{\frac{1}{3}}^1 {{{\left( {\frac{1}{{{x^2}}} - 1} \right)}^{\frac{1}{3}}}\frac{1}{{{x^2}}}.\frac{{dx}}{x}} \)
Đặt : \(t = \left( {\frac{1}{{{x^2}}} - 1} \right) \Rightarrow t + 1 = \frac{1}{{{x^2}}} \Leftrightarrow \left\{ \begin{array}{l}dt = - \frac{{dx}}{x}\\x = \frac{1}{3} \to t = 8;x = 1 \to t = 0\end{array} \right.\)
Khi đó \(I = - \int\limits_8^0 {{t^{\frac{1}{3}}}\left( {t + 1} \right)dt} = \int\limits_0^8 {\left( {{t^{\frac{4}{3}}} + {t^{\frac{1}{3}}}} \right)dt} = \left( {\frac{3}{7}{t^{\frac{7}{3}}} + \frac{3}{4}{t^{\frac{4}{3}}}} \right)\left| {\begin{array}{*{20}{c}}8\\0\end{array} = \frac{3}{7}{{.2}^7} + \frac{3}{4}{{.2}^4} = 16\left( {\frac{{24}}{7} + \frac{3}{4}} \right) = \frac{{468}}{7}} \right.\)
* Chú ý : Còn có cách khác
Vì : \(x \in \left[ {\frac{1}{3};1} \right] \to x \ne 0\).
Đặt \(x = \frac{1}{t} \Rightarrow dx = - \frac{1}{{{t^2}}}dt;f(x)dx = \frac{{{{\left( {\frac{1}{t} - \frac{1}{{{t^3}}}} \right)}^{\frac{1}{3}}}}}{{{{\left( {\frac{1}{t}} \right)}^4}}}\left( { - \frac{1}{{{t^2}}}} \right)dt = - \frac{{{t^2}{{\left( {{t^3} - t} \right)}^{\frac{1}{3}}}}}{t}dt\) \( = - t{\left( {{t^3} - t} \right)^{\frac{1}{3}}}dt = dt = - {t^2}{\left( {1 - \frac{1}{{{t^2}}}} \right)^{\frac{1}{3}}}dt\)(2) .
Đặt : \(u = 1 - \frac{1}{{{t^2}}} \Leftrightarrow \frac{1}{{{t^2}}} = 1 - u;du = \frac{1}{t}dt\)

Ví dụ 3. Tính các tích phân sau:
a)\(\int\limits_1^{{e^{\frac{1}{{p + 2}}}}} {\frac{{{x^{\frac{p}{2}}}}}{{{x^{p + 2}} + 1}}dx} \)
b) \(\int\limits_0^a {\frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}} \)
c) \(\int\limits_0^1 {{e^{x + {e^x}}}dx} \) d. \(\int\limits_0^{2a} {x\sqrt {2ax - {x^2}} dx} \)

Giải

a) \(\int\limits_1^{{e^{\frac{1}{{p + 2}}}}} {\frac{{{x^{\frac{p}{2}}}}}{{{x^{p + 2}} + 1}}dx} \)
Ta có : \(f(x)dx = \frac{{{x^{\frac{p}{2}}}dx}}{{{{\left( {{x^{\frac{{p + 2}}{2}}}} \right)}^2} + 1}}\).
- Đặt : \(t = {x^{\frac{{p + 2}}{2}}} = {x^{\frac{p}{2} + 1}} \Rightarrow \left[ \begin{array}{l}dt = {x^{\frac{p}{2}}}dx\\x = 1 \to t = 1;x = {e^{\frac{1}{{p + 2}}}} \to t = \sqrt e \end{array} \right. \Leftrightarrow I = \int\limits_1^{\sqrt e } {\frac{{dt}}{{{t^2} + 1}}} \)
- Đặt : \(t = \tan u \Rightarrow \left[ \begin{array}{l}dt = \frac{{du}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}u}}\\t = 1 \to u = \frac{\pi }{4},t = {e^{\frac{1}{2}}} \to u = {u_1}\end{array} \right. \Leftrightarrow I = \int\limits_{\frac{\pi }{4}}^{{u_1}} {\frac{{du}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}u\left( {1 + {{\tan }^2}u} \right)}} = \int\limits_{\frac{\pi }{4}}^{{u_1}} {du = \frac{\pi }{4} - {u_1}} } \)
- Từ : \(\tan u = \sqrt e \Rightarrow u = {u_1} = {\rm{artan}}\sqrt {\rm{e}} \Leftrightarrow I = \frac{\pi }{4} - {\rm{artan}}\sqrt {\rm{e}} \)

b) \(\int\limits_0^a {\frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}} \).
Đặt : \(x = {\rm{atant}} \Rightarrow \left\{ \begin{array}{l}{\rm{dx = a}}\frac{{{\rm{dt}}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}t}};x = 0 \to t = 0,x = a \to t = \frac{\pi }{4}\\f(x) = \frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{a^3}{{\tan }^3}t}}{{{a^3}{{\left( {\frac{1}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}} \right)}^{\frac{3}{2}}}}}a\frac{{{\rm{dt}}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}t}} = a\cos t.{\tan ^3}tdt\end{array} \right.\) Vậy : \(I = \int\limits_0^a {f(x)dx} = \int\limits_0^{\frac{\pi }{4}} {a\cos t.{{\tan }^3}tdt} = \int\limits_0^{\frac{\pi }{4}} {a\cos t.\frac{{{{\sin }^3}t}}{{c{\rm{o}}{{\rm{s}}^{\rm{3}}}t}}dt} = \int\limits_0^{\frac{\pi }{4}} {a.\frac{{{{\sin }^3}t}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt = } a\int\limits_0^{\frac{\pi }{4}} {\frac{{\left( {1 - c{\rm{o}}{{\rm{s}}^{\rm{2}}}t} \right)\sin t}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}t}}dt = } \)
- Đặt : \(c{\rm{ost = u}} \Rightarrow \left\{ \begin{array}{l}du = - {\mathop{\rm s}\nolimits} {\rm{intdt;t = }}\frac{\pi }{4} \to u = \frac{1}{{\sqrt 2 }};t = 0 \to u = 1\\f(t)dt = \frac{{\left( {1 - {u^2}} \right)}}{{{u^2}}}\left( { - du} \right) = \left( {1 - \frac{1}{{{u^2}}}} \right)du\end{array} \right.\)
Vậy : \(I = \int\limits_1^{\frac{{\sqrt 2 }}{2}} {\left( {1 - \frac{1}{{{u^2}}}} \right)du = \left( {u + \frac{1}{u}} \right)\left| {\begin{array}{*{20}{c}}{\frac{{\sqrt 2 }}{2}}\\1\end{array} = \frac{{\sqrt 2 }}{2} + \frac{2}{{\sqrt 2 }} - 2 = \frac{3}{{\sqrt 2 }} - 2 = \frac{{3\sqrt 2 }}{2} - 2 = \frac{{3\sqrt 2 - 4}}{2}} \right.} \)

c) \(\int\limits_0^1 {{e^{x + {e^x}}}dx} = \int\limits_0^1 {{e^x}{e^{{e^x}}}dx} \)
Đặt : \(t = {e^x} \Rightarrow \left\{ \begin{array}{l}dt = {e^x}dx;x = 0 \to t = 1;x = 1 \to t = e\\f(x)dx = {e^x}{e^{{e^x}}}dx = {e^t}dt\end{array} \right.\)
Vậy : \(I = \int\limits_0^1 {f(x)dx} = \int\limits_1^e {{e^t}dt} = {e^t}\left| {\begin{array}{*{20}{c}}e\\1\end{array} = {e^e} - e} \right.\)

d) \(\int\limits_0^{2a} {x\sqrt {2ax - {x^2}} dx} = \int\limits_0^{2a} {x\sqrt {{a^2} - {{\left( {x - a} \right)}^2}} dx} \)
Đặt : \(x - a = a.\sin t \Rightarrow \left\{ \begin{array}{l}dx = a.c{\rm{ostdt,x = 0}} \to {\rm{t = - }}\frac{\pi }{2}{\rm{;x = 2a}} \to {\rm{t = }}\frac{\pi }{2}\\f(x)dx = \left( {a + a.\sin t} \right)\sqrt {{a^2}c{\rm{o}}{{\rm{s}}^{\rm{2}}}t} .a.c{\rm{ostdt}}\end{array} \right.\)
Vậy: \(I = {a^3}\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + \sin t} \right)c{\rm{o}}{{\rm{s}}^{\rm{2}}}tdt} = {a^3}\left[ {\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\rm{o}}{{\rm{s}}^{\rm{2}}}tdt + \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\rm{o}}{{\rm{s}}^{\rm{2}}}t\sin tdt} } } \right] = {a^3}\left[ {\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{1 + c{\rm{os2}}t}}{2}dt - \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\rm{o}}{{\rm{s}}^{\rm{2}}}td\left( {c{\rm{os}}t} \right)} } } \right]\) \( = {a^3}\left[ {\frac{1}{2}\left( {t + \frac{1}{2}\sin 2t} \right)\left| {\begin{array}{*{20}{c}}{\frac{\pi }{2}}\\{ - \frac{\pi }{2}}\end{array} - \frac{1}{3}{\rm{co}}{{\rm{s}}^{\rm{3}}}t\left| {\begin{array}{*{20}{c}}{\frac{\pi }{2}}\\{ - \frac{\pi }{2}}\end{array}} \right.} \right.} \right] = {a^3}\left[ {\frac{1}{2}\left( {\frac{\pi }{2} + \frac{\pi }{2}} \right)} \right] = {a^3}\frac{\pi }{2}\)

Ví dụ 4. Tính các tích phân sau
a) \(\int\limits_2^3 {\frac{{dx}}{{{x^5} - {x^2}}}} \)
b) \(\int\limits_0^1 {\frac{{{x^7}dx}}{{{{\left( {1 + {x^4}} \right)}^2}}}} \)
c) \(\int\limits_0^1 {\frac{{{x^3} - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} \)
d) \(\int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^4}}}dx} \)

Giải

a) \(\int\limits_2^3 {\frac{{dx}}{{{x^5} - {x^2}}}} = \int\limits_2^3 {\frac{1}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx\quad \left( 1 \right)} \)
Xét : \(f(x) = \frac{1}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{A}{{{x^2}}} + \frac{B}{x} + \frac{{Cx + D}}{{{x^2} + x + 1}} + \frac{E}{{x - 1}}\)
\( = \frac{{A\left( {{x^2} + x + 1} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {Cx + D} \right){x^2}\left( {x - 1} \right) + E({x^2} + x + 1){x^2}}}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\)
\( = \frac{{\left( {B + C + E} \right){x^4} + \left( {A + D - C + E} \right){x^3} + \left( {E - D} \right){x^2} - Bx - A}}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\).
Đồng nhất hệ số hai tử số ta có hệ:
\(\left\{ \begin{array}{l}B + C + E = 0\\A + D - C + E = 0\\E - D = 0\\B = 0\\A = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}C = - E\\E + E + E = 1\\B = 0\\E = D\\A = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}D = \frac{1}{3}\\C = - \frac{1}{3}\\B = 0\\E = \frac{1}{3}\\A = - 1\end{array} \right. \Rightarrow f(x) = - \frac{1}{{{x^2}}} + \frac{{ - \frac{1}{3}x + \frac{1}{3}}}{{{x^2} + x + 1}} + \frac{{\frac{1}{3}}}{{x - 1}}\)
Vậy : \(I = \int\limits_2^3 {\left( { - \frac{1}{{{x^2}}} + \frac{{ - \frac{1}{3}x + \frac{1}{3}}}{{{x^2} + x + 1}} + \frac{{\frac{1}{3}}}{{x - 1}}} \right)dx} = \int\limits_2^3 {\left( { - \frac{1}{{{x^2}}} - \frac{1}{3}\left( {\frac{{x - 1}}{{{x^2} + x + 1}}} \right) + \frac{1}{3}\frac{1}{{\left( {x - 1} \right)}}} \right)dx} \) \( = \left( {\frac{1}{x} - \frac{1}{6}\ln \left| {{x^2} + x + 1} \right| + \frac{1}{3}\ln \left| {x - 1} \right|} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array} - \int\limits_2^3 {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} } \right. = \left( {\frac{1}{x} + \frac{1}{6}\ln \frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} + \frac{1}{{\sqrt 3 }}{\rm{arctan}}\frac{{{\rm{2x + 1}}}}{{\sqrt 3 }}} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array}} \right.\)\( = \frac{1}{6} + \frac{1}{{\sqrt 3 }}\left( {{\rm{arctan}}\frac{{\rm{7}}}{{\sqrt {\rm{3}} }} - {\rm{arctan}}\frac{{\rm{5}}}{{\sqrt {\rm{3}} }}} \right)\)

b) \(\int\limits_0^1 {\frac{{{x^7}dx}}{{{{\left( {1 + {x^4}} \right)}^2}}}} = \frac{1}{3}\int\limits_0^1 {\frac{{{x^4}}}{{{{\left( {1 + {x^4}} \right)}^2}}}3{x^3}dx\quad \left( 1 \right)} \).
Đặt : \(t = 1 + {x^4} \Rightarrow \left\{ \begin{array}{l}dt = 3{x^3}dx,x = 0 \to t = 1;x = 1 \to t = 2\\f(x)dx = \frac{1}{3}\left( {\frac{{t - 1}}{{{t^2}}}} \right)dt = \frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)dt\end{array} \right.\)
Vậy : \(I = \int\limits_0^2 {\frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)dt = \frac{1}{3}\left( {\ln \left| t \right| + \frac{1}{t}} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{1}{3}\left( {\ln 2 - \frac{1}{2}} \right)} \right.} \)
c) \(\int\limits_0^1 {\frac{{{x^3} - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} = \frac{1}{2}\int\limits_0^1 {\frac{{\left( {{x^2} - 2} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}2xdx} \quad \left( 1 \right)\)
Đặt : \(t = 1 + {x^2} \Leftrightarrow {x^2} - 2 = t - 3 \Rightarrow \left\{ \begin{array}{l}dt = 2xdx;x = 0 \to t = 1;x = 1 \to t = 2\\f(x)dx = \frac{1}{2}\left( {\frac{{t - 3}}{{{t^2}}}} \right)dt = \frac{1}{2}\left( {\frac{1}{t} - \frac{3}{{{t^2}}}} \right)dt\end{array} \right.\)
Vậy : \(I = \int\limits_1^2 {\frac{1}{2}\left( {\frac{1}{t} - \frac{3}{{{t^2}}}} \right)dt = \frac{1}{2}\left( {\ln \left| t \right| + \frac{3}{t}} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{1}{2}\left( {\ln 2 - \frac{3}{2}} \right)} \right.} \)
d) \(\int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^4}}}dx} = \int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^6}}}{x^2}dx} \quad \left( 1 \right)\).
Đặt : \(t = \sqrt {1 + {x^3}} \leftrightarrow {t^2} = 1 + {x^3} \leftrightarrow \left\{ \begin{array}{l}2tdt = 3{x^2}dx;x = 1 \to t = \sqrt 2 ,x = 2 \to t = 3\\f(x)dx = \frac{1}{3}\frac{{\sqrt {1 + {x^3}} }}{{{x^6}}}3{x^2}dx = \frac{1}{3}\frac{t}{{{{\left( {{t^2} - 1} \right)}^2}}}2tdt = \frac{2}{3}\frac{{{t^2}}}{{{{\left( {{t^2} - 1} \right)}^2}}}dt\end{array} \right.\)
Vậy : \(I = \frac{2}{3}\int\limits_{\sqrt 2 }^3 {{{\left( {\frac{1}{{t + 1}} + \frac{1}{2}\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)} \right)}^2}dt = \frac{2}{3}\left[ {\int\limits_{\sqrt 2 }^3 {\frac{1}{4}{{\left( {\frac{1}{{t + 1}} - \frac{1}{{t - 1}}} \right)}^2}} } \right] = \frac{1}{6}\int\limits_{\sqrt 2 }^3 {\left( {\frac{1}{{{{\left( {t + 1} \right)}^2}}} + \frac{1}{{{{\left( {t - 1} \right)}^2}}} - \left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)} \right)dt} } \)\( = \frac{1}{6}\left[ { - \frac{1}{{t + 1}} - \frac{1}{{t - 1}} - \ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right]\left| {\begin{array}{*{20}{c}}3\\{\sqrt 2 }\end{array} = \frac{1}{6}\left( {\frac{{ - 2t}}{{\left( {{t^2} - 1} \right)}} - \ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| {\begin{array}{*{20}{c}}3\\{\sqrt 2 }\end{array}} \right.} \right. = \frac{{8\sqrt 2 - 3}}{{24}} + \frac{1}{3}\ln \left( {2\sqrt 2 - 2} \right).\)

Ví dụ 5. Tính các tích phân sau :
a) \(\int\limits_{\sqrt 7 }^4 {\frac{{dx}}{{x\sqrt {{x^2} + 9} }}} \)
b) \(\int\limits_0^1 {\frac{{\left( {{x^2} - x} \right)dx}}{{\sqrt {{x^2} + 1} }}} \)
c) \(\int\limits_0^{\sqrt 3 } {\frac{{{x^5} - 2{x^3}}}{{\sqrt {{x^2} + 1} }}dx} \)
d) \(\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx} \)

Giải

a.) \(\int\limits_{\sqrt 7 }^4 {\frac{{dx}}{{x\sqrt {{x^2} + 9} }}} = \int\limits_{\sqrt 7 }^4 {\frac{{xdx}}{{{x^2}\sqrt {{x^2} + 9} }}} \quad \left( 1 \right)\).
Đặt : \(t = \sqrt {{x^2} + 9} \Rightarrow \left\{ \begin{array}{l}{t^2} = {x^2} + 9 \leftrightarrow tdt = xdx,{x^2} = {t^2} - 9\\x = \sqrt 7 \to t = 4,x = 4 \to t = 5\end{array} \right.\).
Do đó : \(I = \int\limits_4^5 {\frac{{dt}}{{t\left( {{t^2} - 9} \right)}} = } \int\limits_4^5 {\frac{{dt}}{{t\left( {t - 3} \right)\left( {t + 3} \right)}}} \)
Ta có : \(f(t) = \frac{1}{{t\left( {t - 3} \right)\left( {t + 3} \right)}} = \frac{A}{t} + \frac{B}{{t - 3}} + \frac{C}{{t + 3}} = \frac{{A\left( {{t^2} - 9} \right) + Bt\left( {t + 3} \right) + C\left( {t - 3} \right)t}}{{t\left( {{t^2} - 9} \right)}}\)
Đồng nhất hệ số hai tử số bằng cách thay lần lượt các nghiệm vào hai tử số ta có:
- Với x=0 : -9A=1 \( \to A = - \frac{1}{9}\)
- Với x=-3 : 9C=1 \( \to C = \frac{1}{9}\)
- Với x=3 : 9B=1 \( \to B = \frac{1}{9}\)
Vậy : \(I = \frac{1}{9}\left[ {\int\limits_4^5 {\left( { - \frac{1}{t} + \frac{1}{{t - 3}} + \frac{1}{{t + 3}}} \right)dt} } \right] = \frac{1}{9}\left[ {\ln \left( {{t^2} - 9} \right) - \ln t} \right]\left| {\begin{array}{*{20}{c}}5\\4\end{array} = \frac{1}{9}\ln \frac{{{t^2} - 9}}{t}\left| {\begin{array}{*{20}{c}}5\\4\end{array} = \frac{1}{9}\ln \frac{{144}}{{35}}} \right.} \right.\)
* Chú ý : Nếu theo phương pháp chung thì đặt : \(x = 3\sin t \to dx = 3\cos tdt\).
Khi : \(\left\{ \begin{array}{l}x = \sqrt 7 \to \sqrt 7 = 3\sin t \leftrightarrow \sin t = \frac{{\sqrt 7 }}{3}\\x = 4 \to 4 = 3\sin t \leftrightarrow \sin t = \frac{4}{3} > 1\end{array} \right.\). Như vậy ta không sử dụng được phương pháp này được.

b) \(\int\limits_0^1 {\frac{{\left( {{x^2} - x} \right)dx}}{{\sqrt {{x^2} + 1} }}} = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}dx - \int\limits_0^1 {\frac{x}{{\sqrt {{x^2} + 1} }}dx} = J - K\quad \left( 1 \right)} \)
* Để tính \(J\):
Đặt : \(x = \tan t \Rightarrow \left\{ \begin{array}{l}dx = \frac{1}{{c{\rm{o}}{{\rm{s}}^2}t}}dt,x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}\\f(x)dx = \frac{{{{\tan }^2}t.\frac{1}{{c{\rm{o}}{{\rm{s}}^2}t}}dt}}{{\sqrt {1 + {{\tan }^2}t} }} = \frac{{{{\tan }^2}t}}{{c{\rm{ost}}}}dt\end{array} \right.\).
Tính tích phân này không đơn giản , vì vậy ta phải có cách khác . - Từ : \(g(x) = \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }} = \frac{{{x^2} + 1 - 1}}{{\sqrt {{x^2} + 1} }} = \sqrt {{x^2} + 1} - \frac{1}{{\sqrt {{x^2} + 1} }} \Rightarrow \int\limits_0^1 {g(x)dx = \int\limits_0^1 {\sqrt {{x^2} + 1} dx - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx} } } \)
- Hai tích phân này đều tính được .
+/ Tính : \(E = \int\limits_0^1 {\sqrt {{x^2} + 1} dx = } x\sqrt {{x^2} + 1} \left| {\begin{array}{*{20}{c}}1\\0\end{array} - \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}dx = } } \right.\sqrt 2 - \left( {\int\limits_0^1 {\sqrt {{x^2} + 1} } dx - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx} } \right)\)
\( = \sqrt 2 - E + \ln \left| {x + \sqrt {{x^2} + 1} } \right|\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. \Rightarrow 2E = \sqrt 2 + \ln \left( {1 + \sqrt 2 } \right) \Leftrightarrow E = \frac{{\sqrt 2 }}{2} + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right)\)
* Tính \(K=\int\limits_0^1 {\frac{x}{{\sqrt {{x^2} + 1} }}dx = \sqrt {{x^2} + 1} \left| {\begin{array}{*{20}{c}}1\\0\end{array} = \sqrt 2 - 1} \right.} \); \(\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx = \ln \left| {x + \sqrt {{x^2} + 1} } \right|\left| {\begin{array}{*{20}{c}}1\\0\end{array} = \ln \left( {1 + \sqrt 2 } \right)} \right.} \)
Do đó: \(I=\frac{{\sqrt 2 }}{2} + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right) + \ln \left( {1 + \sqrt 2 } \right) = \frac{{\sqrt 2 }}{2} + \frac{3}{2}\ln \left( {1 + \sqrt 2 } \right)\)

c) \(\int\limits_0^{\sqrt 3 } {\frac{{{x^5} - 2{x^3}}}{{\sqrt {{x^2} + 1} }}dx} = \int\limits_0^{\sqrt 3 } {\frac{{{x^5}}}{{\sqrt {{x^2} + 1} }}dx - 2\int\limits_0^{\sqrt 3 } {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx = J - K\left( 1 \right)} } \)
- Tính \(J\): Đặt \(t = \sqrt {{x^2} + 1} \Rightarrow \left\{ \begin{array}{l}{x^2} = {t^2} - 1;xdx = tdt;x = 0 \to t = 1,x = \sqrt 3 \to t = 2\\f(x)dx = \frac{{{x^4}xdx}}{{\sqrt {{x^2} + 1} }} = \frac{{{{\left( {{t^2} - 1} \right)}^2}tdt}}{t} = \left( {{t^4} - 2{t^2} + 1} \right)dt\end{array} \right.\)
Suy ra : \(J=\int\limits_1^2 {\left( {{t^4} - 2{t^2} + 1} \right)dt = \left( {\frac{1}{5}{t^5} - \frac{2}{3}{t^3} + t} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{{38}}{{15}}} \right.} \)
- Tính \(K\): Đặt \(t = \sqrt {{x^2} + 1} \Rightarrow \left\{ \begin{array}{l}{x^2} = {t^2} - 1;xdx = tdt;x = 0 \to t = 1,x = \sqrt 3 \to t = 2\\f(x)dx = \frac{{{x^2}xdx}}{{\sqrt {{x^2} + 1} }} = \frac{{\left( {{t^2} - 1} \right)tdt}}{t} = \left( {{t^2} - 1} \right)dt\end{array} \right.\)
Suy ra: \(K=\int\limits_1^2 {\left( {{t^2} - 1} \right)dt = \left( {\frac{1}{3}{t^3} - t} \right)\left| {\begin{array}{*{20}{c}}2\\1\end{array} = \frac{4}{3}} \right.} \)
Vậy : I=\(\frac{{28}}{{15}} + \frac{4}{3} = \frac{{48}}{{15}} = \frac{{16}}{5}\)

d) \(\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx} \).
Đặt : \(x = \sin t \to \left\{ \begin{array}{l}dx = c{\rm{ostdt}}{\rm{. x = 0}} \to {\rm{t = 0;x = 1}} \to {\rm{t = }}\frac{\pi }{2}\\f(x)dx = \sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx = \sqrt {c{\rm{o}}{{\rm{s}}^{\rm{6}}}t} c{\rm{ostdt = co}}{{\rm{s}}^{\rm{4}}}tdt\end{array} \right.\)
Do đó I=\(\int\limits_0^{\frac{\pi }{2}} {{{\left( {\frac{{1 - c{\rm{os2t}}}}{2}} \right)}^2}dt = \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - 2\cos 2t + \frac{{1 + c{\rm{os4t}}}}{2}} \right)dt = \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{3}{4} - \frac{1}{2}c{\rm{os2t + }}\frac{1}{{\rm{8}}}c{\rm{os4t}}} \right)dt} } } \)
\( = \left( {\frac{3}{4}t - \frac{1}{4}\sin 2t + \frac{1}{{32}}\sin 4t} \right)\left| {\begin{array}{*{20}{c}}{\frac{\pi }{2}}\\0\end{array} = \frac{{3\pi }}{8}} \right.\)

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