Question

问一道不定积分的题目求（arctanx)^2的原函数

Answer 1

用分部积分法

∫(arctanx)^2dx

=x(arctanx)^2-∫[x*2arctanx*(1/1+x^2)]dx

=x(arctanx)^2-∫[2x/(1+x^2)]arctanxdx

再对∫[2x/(1+x^2)]arctanxdx先换元,再分部积分：

令u=arctanx,du=1/(1+x^2)

x=tanu,2x=2tanu

∫[2x/(1+x^2)]arctanxdx

=∫2utanudu

=∫tanud(u^2)

=u^2tanu-∫u^2/(1+u^2)du

=u^2tanu-∫(u^2+1-1)/(u^2+1)du

=u^2tanu-∫[1-1/(1+u^2)]du

=u^2tanu-u+arctanu

=x(arctanx)^2-arctanx+tanx+C

原式

=x(arctanx)^2-x(arctanx)^2+arctanx-tanx+C