请问,这个不定积分怎么求?
发布网友
发布时间：2023-12-04 09:43
共1个回答
热心网友
时间：2024-12-05 07:47
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))]
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))] = (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。😊
