（1）分解为部分分式。设

$$\frac{1}{(x+1)(x^2-2x+2)}=\frac{A}{x+1}+\frac{Bx+C}{x^2-2x+2}.$$

两边同乘分母并比较系数，得 $A+B=0$、$-2A+B+C=0$、$2A+C=1$，解得 $A=\frac{1}{5}$、$B=-\frac{1}{5}$、$C=\frac{3}{5}$。因此所求积分为

$$I=\frac{1}{5}{\int }_0^1\frac{\mathrm{d}x}{x+1}+\frac{1}{5}{\int }_0^1\frac{-x+3}{x^2-2x+2}\mathrm{d}x.$$

（2）第一项为 $\frac{\ln 2}{5}$。第二项利用 $-x+3=-\frac{1}{2}(2x-2)+2$、$x^2-2x+2=(x-1{)}^2+1$，得

$$\begin{array}{rl}{\int }_0^1\frac{-x+3}{x^2-2x+2}\mathrm{d}x& ={\left[-\frac{1}{2}\ln (x^2-2x+2)+2\arctan (x-1)\right]}_0^1\\ & =\frac{1}{2}\ln 2+\frac{\pi }{2}.\end{array}$$

（3）代回得 $I=\frac{1}{5}\ln 2+\frac{1}{5}(\frac{1}{2}\ln 2+\frac{\pi }{2})=\frac{3\ln 2+\pi }{10}$。