# 部分分式积分法

## 例子

${\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={10x^{2}+12x+20 \over (x-2)(x^{2}+2x+4)}={A \over x-2}+{Bx+C \over x^{2}+2x+4}}$

${\displaystyle 10x^{2}+12x+20=A(x^{2}+2x+4)+(Bx+C)(x-2)\,}$

${\displaystyle 10x^{2}+12x+20=(A+B)x^{2}+(2A-2B+C)x+(4A-2C)\,}$

${\displaystyle A+B=10\,}$
${\displaystyle 2A-2B+C=12\,}$
${\displaystyle 4A-2C=20\,}$

${\displaystyle A=7\,}$
${\displaystyle B=3\,}$
${\displaystyle C=4\,}$

${\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={7 \over x-2}+{3x+4 \over x^{2}+2x+4}}$

${\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx=\int ({7 \over x-2}+{3x+4 \over x^{2}+2x+4})\,dx=\int {7 \over x-2}\,dx+\int {3x+4 \over x^{2}+2x+4}\,dx}$

${\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx}$
${\displaystyle =7\ln |x-2|+\int {{{\frac {3}{2}}(2x+2)+1} \over x^{2}+2x+4}\,dx}$
${\displaystyle =7\ln |x-2|+{\frac {3}{2}}\int {2x+2 \over x^{2}+2x+4}\,dx+\int {1 \over (x+1)^{2}+3}\,dx}$
${\displaystyle =7\ln |x-2|+{\frac {3}{2}}\ln |x^{2}+2x+4|+{\frac {1}{\sqrt {3}}}\arctan({x+1 \over {\sqrt {3}}})+C}$