题干
$$ \int_0^1 \frac{1}{(x+1)(x^2-2x+2)}dx. $$解题
设
$$ \frac{1}{(x+1)(x^2-2x+2)}=\frac{A}{x+1}+\frac{Bx+D}{x^2-2x+2} $$通分
$$ (A+B)x^2+(B+D-2A)x+2A+D-1=0 $$列方程组
$$ \left \{ \begin{aligned} & A+B=0 \\ & D+B-2A =0 \\ & 2A+D-1 =0 \end{aligned} \right. $$解得
$$ \left \{ \begin{aligned} & A=\frac{1}{5} \\ & B=-\frac{1}{5} \\ & D=\frac{3}{5} \end{aligned} \right. $$代入
$$ \frac{1}{(x+1)(x^2-2x+2)}=\frac{1}{5(x+1)}+\frac{-x+3}{5(x^2-2x+2)} $$剩下的就是有理函数的积分的内容了