Question

求不定积分[x^2+1/(x+1)^2(x-1)]dx

Answer 1

这个不定积分的解法是：

• 将被积函数分解为部分分式的形式，即

[x2+1/(x+1)2(x-1)]dx = [x2/(x+1)2(x-1)]dx + [1/(x+1)^2(x-1)]dx

• 对第一项，使用换元法，令u = x+1，得到

[x2/(x+1)2(x-1)]dx = [(u-1)2/(u2(u-2))]du

• 对第二项，使用分部积分法，令u = 1/(x+1)，v’ = 1/(x-1)，得到

[1/(x+1)^2(x-1)]dx = [u/(x-1)]dx - [u’/(x-1)]dx

• 分别对两项求不定积分，得到

∫[(u-1)2/(u2(u-2))]du = (1/2)ln|u-2| - (1/2)ln|u| - (1/u) + C

∫[u/(x-1)]dx = (1/2)ln|x-1| + C

∫[u’/(x-1)]dx = (1/2)ln|x-1| - (1/x) + C

• 将u替换回x+1，合并常数项，得到最终答案

∫[x2+1/(x+1)2(x-1)]dx = (1/2)ln|x-1| - (1/2)ln|x+1| - (1/(x+1)) - (1/x) + C

希望这个答案对您有帮助。😊