微積分/積分技巧/部分分式分解/解答

使用部分分式分解方法計算以下積分。

1. $\int {\frac {2x+11}{(x+6)(x+5)}}dx$

分解分式

${\frac {2x+11}{(x+6)(x+5)}}={\frac {A}{x+6}}+{\frac {B}{x+5}}={\frac {Ax+5A+Bx+6B}{(x+6)(x+5)}}$

將 x 的係數相等

$A+B=2$

$5A+6B=11$

求解方程組

$A=1,B=1$

改寫積分並求解

${\begin{aligned}\int {\frac {2x+11}{(x+6)(x+5)}}dx&=\int {\frac {dx}{x+6}}+\int {\frac {dx}{x+5}}\\&=\mathbf {\ln |x+6|+\ln |x+5|+C} \end{aligned}}$

2. $\int {\frac {7x^{2}-5x+6}{(x-1)(x-3)(x-7)}}dx$

分解分式

${\begin{aligned}{\frac {7x^{2}-5x+6}{(x-1)(x-3)(x-7)}}&={\frac {A}{x-1}}+{\frac {B}{x-3}}+{\frac {C}{x-7}}\\&={\frac {A(x-3)(x-7)+B(x-1)(x-7)+C(x-1)(x-3)}{(x-1)(x-3)(x-7)}}\\&={\frac {A(x^{2}-10x+21)+B(x^{2}-8x+7)+C(x^{2}-4x+3)}{(x-1)(x-3)(x-7)}}\\&={\frac {Ax^{2}-10Ax+21A+Bx^{2}-8Bx+7B+Cx^{2}-4Cx+3C}{(x-1)(x-3)(x-7)}}\end{aligned}}$

係數相等

${\begin{aligned}A+B+C&=7\\-10A-8B-4C&=-5\\21A+7B+3C&=6\end{aligned}}$

求解方程組

$D={\Biggr |}{\begin{array}{ccc}1&1&1\\-10&-8&-4\\21&7&3\end{array}}{\Biggr |}=-24-84-70-(-30-28-168)=48$

$A={\frac {{\Biggr |}{\begin{array}{ccc}7&1&1\\-5&-8&-4\\6&7&3\end{array}}{\Biggr |}}{D}}={\frac {(7)(-8)(3)-24-35-(-15+(7)(-4)(7)-48)}{48}}={\frac {32}{48}}={\frac {2}{3}}$

$B={\frac {{\Biggr |}{\begin{array}{ccc}1&7&1\\-10&-5&-4\\21&6&3\end{array}}{\Biggr |}}{D}}={\frac {-15+(7)(-4)(21)-60-(-210-24+(-5)(21))}{48}}={\frac {-324}{48}}=-{\frac {27}{4}}$

$C={\frac {{\Biggr |}{\begin{array}{ccc}1&1&7\\-10&-8&-5\\21&7&6\end{array}}{\Biggr |}}{D}}={\frac {(-8)(6)+(-5)(21)+(7)(-10)(7)-((-10)(6)+(-5)(7)+(7)(-8)(21))}{48}}={\frac {628}{48}}={\frac {157}{12}}$

改寫積分並求解

${\begin{aligned}\int {\frac {7x^{2}-5x+6}{(x-1)(x-3)(x-7)}}dx&={\frac {2}{3}}\int {\frac {dx}{x-1}}-{\frac {27}{4}}\int {\frac {dx}{x-3}}+{\frac {157}{12}}\int {\frac {dx}{x-7}}\\&=\mathbf {{\frac {2}{3}}\ln |x-1|-{\frac {27}{4}}\ln |x-3|+{\frac {157}{12}}\ln |x-7|+C} \end{aligned}}$

3. $\int {\frac {x^{2}-x+2}{x(x+2)^{2}}}dx$

分解分式

${\begin{aligned}{\frac {x^{2}-x+2}{x(x+2)^{2}}}&={\frac {A}{x}}+{\frac {B}{x+2}}+{\frac {C}{(x+2)^{2}}}\\&={\frac {A(x+2)^{2}+Bx(x+2)+Cx}{x(x+2)^{2}}}\\&={\frac {A(x^{2}+4x+4)+Bx^{2}+2Bx+Cx}{x(x+2)^{2}}}\\&={\frac {Ax^{2}+4Ax+4A+Bx^{2}+2Bx+Cx}{x(x+2)^{2}}}\end{aligned}}$

將係數等同

${\begin{array}{ccccccc}A&+&B&&&=&1\\4A&+&2B&+&C&=&-1\\4A&&&&&=&2\end{array}}$

求解方程組

$4A=2\implies A={\frac {1}{2}}$

$A+B=1\implies B={\frac {1}{2}}$

$4A+2B+C=-1\implies C=-4$

改寫積分並求解

${\begin{aligned}\int {\frac {x^{2}-x+2}{x(x+2)^{2}}}dx&={\frac {1}{2}}\int {\frac {dx}{x}}+{\frac {1}{2}}\int {\frac {dx}{x+2}}-4\int {\frac {dx}{(x+2)^{2}}}\\&=\mathbf {{\frac {1}{2}}\ln |x|+{\frac {1}{2}}\ln |x+2|+{\frac {4}{x+2}}+C} \end{aligned}}$

4. $\int {\frac {2}{(x+2)(x^{2}+3)}}dx$

:${\frac {2}{(x+2)(x^{2}+3)}}={\frac {A}{x+2}}+{\frac {Bx+C}{x^{2}+3}}={\frac {Ax^{2}+3A+Bx^{2}+Cx+2Bx+2C}{(x+2)(x^{2}+3)}}$

將每個x次冪的係數相等。對於$x^{2}$

$A+B=0$

, 對於$x$

$C+2B=0$

, 以及常數項

$3A+2C=2$

用你喜歡的方式求解方程組（這裡使用高斯消元法和回代）

$\left[{\begin{array}{cccc}1&1&0&0\\0&2&1&0\\3&0&2&2\end{array}}\right]\left[{\begin{array}{cccc}1&1&0&0\\0&1&{\frac {1}{2}}&0\\0&-3&2&2\end{array}}\right]\left[{\begin{array}{cccc}1&1&0&0\\0&1&{\frac {1}{2}}&0\\0&0&{\frac {7}{2}}&2\end{array}}\right]\left[{\begin{array}{cccc}1&1&0&0\\0&1&{\frac {1}{2}}&0\\0&0&1&{\frac {4}{7}}\end{array}}\right]$

$C={\frac {4}{7}},\,B=-{\frac {2}{7}},\,A={\frac {2}{7}}$

所以

$\int {\frac {2}{(x+2)(x^{2}+3)}}dx={\frac {2}{7}}\int {\frac {1}{x+2}}dx-{\frac {2}{7}}\int {\frac {x}{x^{2}+3}}+{\frac {4}{7}}\int {\frac {1}{x^{2}+3}}dx$

要計算第一個積分，使用替換法，令 $u=x+2$，$du=dx$.
要計算第二個積分，使用替換法，令 $u=x^{2}+3$，$du=2xdx$，$dx={\frac {du}{2x}}$.
要計算第三個積分，使用三角替換法，令 $x={\sqrt {3}}\tan(\theta )$，$dx={\sqrt {3}}\sec ^{2}(\theta )d\theta$.

5. $\int {\frac {dx}{(x+2)(x^{2}+2)}}$

分解分式

${\begin{aligned}{\frac {1}{(x+2)(x^{2}+2)}}&={\frac {A}{x+2}}+{\frac {Bx+C}{x^{2}+2}}\\&={\frac {A(x^{2}+2)+(Bx+C)(x+2)}{(x+2)(x^{2}+2)}}\\&={\frac {Ax^{2}+2A+Bx^{2}+2Bx+Cx+2C}{(x+2)(x^{2}+2)}}\end{aligned}}$

將係數等同

${\begin{array}{ccccccc}A&+&B&&&=&0\\&&2B&+&C&=&0\\2A&&&+&2C&=&1\end{array}}$

求解方程組

$D={\Biggr |}{\begin{array}{ccc}1&1&0\\0&2&1\\2&0&2\end{array}}{\Biggr |}=6$

$A={\frac {{\Biggr |}{\begin{array}{ccc}0&1&0\\0&2&1\\1&0&2\end{array}}{\Biggr |}}{D}}={\frac {1}{6}}$

$B={\frac {{\Biggr |}{\begin{array}{ccc}1&0&0\\0&0&1\\2&1&2\end{array}}{\Biggr |}}{D}}=-{\frac {1}{6}}$

$C={\frac {{\Biggr |}{\begin{array}{ccc}1&1&0\\0&2&0\\2&0&1\end{array}}{\Biggr |}}{D}}={\frac {2}{6}}={\frac {1}{3}}$

改寫積分並求解

$\int {\frac {dx}{(x+2)(x^{2}+2)}}={\frac {1}{6}}\int {\frac {dx}{x+2}}-{\frac {1}{6}}\int {\frac {xdx}{x^{2}+2}}+{\frac {1}{3}}\int {\frac {dx}{x^{2}+2}}$

進行替換

$u=x^{2}+2\qquad du=2xdx\qquad dx={\frac {du}{2x}}$

在第二個積分中，以及

${\frac {x}{\sqrt {2}}}=\tan(\theta )\qquad {\frac {dx}{\sqrt {2}}}=\sec ^{2}(\theta )d\theta \qquad dx={\sqrt {2}}\sec ^{2}(\theta )d\theta$

在第三個積分中，我們有

${\begin{aligned}\int {\frac {dx}{(x+2)(x^{2}+2)}}&={\frac {1}{6}}\ln |x+2|-{\frac {1}{6}}\int {\frac {du}{2u}}+{\frac {1}{3}}\int {\frac {{\sqrt {2}}\sec ^{2}(\theta )d\theta }{2(\tan ^{2}(\theta )+1)}}\\&={\frac {1}{6}}\ln |x+2|-{\frac {1}{12}}\ln |u|+{\frac {\sqrt {2}}{6}}\int {\frac {\sec ^{2}(\theta )d\theta }{\sec ^{2}(\theta )}}\\&={\frac {1}{6}}\ln |x+2|-{\frac {1}{12}}\ln |x^{2}+2|+{\frac {\sqrt {2}}{6}}\int d\theta \\&={\frac {1}{6}}\ln |x+2|-{\frac {1}{12}}\ln |x^{2}+2|+{\frac {\sqrt {2}}{6}}\theta +C\\&=\mathbf {{\frac {1}{6}}\ln |x+2|-{\frac {1}{12}}\ln |x^{2}+2|+{\frac {\sqrt {2}}{6}}\arctan({\frac {x}{\sqrt {2}}})+C} \end{aligned}}$

6. ${\begin{aligned}{\frac {1}{(x^{2}+1)^{2}(x-1)}}&={\frac {Ax+B}{x^{2}+1}}+{\frac {Cx+D}{(x^{2}+1)^{2}}}+{\frac {E}{x-1}}\\&={\frac {(Ax+B)(x^{2}+1)(x-1)+(Cx+D)(x-1)+E(x^{2}+1)^{2}}{(x^{2}+1)^{2}(x-1)}}\\&={\frac {(Ax+B)(x^{3}-x^{2}+x-1)+Cx^{2}+(D-C)x-D+E(x^{4}+2x^{2}+1)}{(x^{2}+1)^{2}(x-1)}}\\&={\frac {Ax^{4}+(B-A)x^{3}+(A-B)x^{2}+(B-A)x-B+Cx^{2}+(D-C)x-D+Ex^{4}+2Ex^{2}+E}{(x^{2}+1)^{2}(x-1)}}\end{aligned}}$