求积分∫(x^2)arctanxdx

Question

求积分∫(x^2)arctanxdx

Answer 1

设u = arctanx,(x^2)dx = v;
那么可以求出v = ∫[x^2/(1+x^2)]dx = x - arctanx;
（分步积分）∫(x^2)arctanxdx= v*arctanx - (v'*arctanx)'
= (x - arctanx)*arctanx - ∫[1-1/(x^2+1)]*[1/(x^2+1)]dx
接下来就好做了
具体思路就是这样的：形如∫(x^n)arctanxdx的题目,令u=arctanx,v=(x^n)dx,然后分部积分.
我顺便复习了下.