4.求不定积分int(x+1)/((x-1)(x^2)+1)^(2)dx.

将被积函数分解为部分分式： \[ \frac{x+1}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \] 通分并比较系数，得： \[ A = \frac{1}{2}, \quad B = -\frac{1}{2}, \quad C = -\frac{1}{2}, \quad D = -1, \quad E = 0 \] 代入得： \[ \frac{x+1}{(x-1)(x^2+1)^2} = \frac{1}{2(x-1)} - \frac{x+1}{2(x^2+1)} - \frac{x}{(x^2+1)^2} \] 分别积分： \[ \int \frac{1}{2(x-1)} \, dx = \frac{1}{2} \ln |x-1| \] \[ \int -\frac{x+1}{2(x^2+1)} \, dx = -\frac{1}{4} \ln (x^2+1) - \frac{1}{2} \arctan x \] \[ \int -\frac{x}{(x^2+1)^2} \, dx = \frac{1}{2(x^2+1)} \] 相加得： \[ \boxed{\frac{1}{2} \ln |x-1| - \frac{1}{4} \ln (x^2+1) - \frac{1}{2} \arctan x + \frac{1}{2(x^2+1)} + C} \]