Question

求不定积分∫arccotxdx 用分部积分法谢谢!

Answer 1

简单计算一下，答案如图所示

Answer 2

原式=xarccotx-∫x*(-1/(1+x^2))dx
=xarccotx+∫x/(1+x^2)dx
=xarccotx+1/2∫d(x^2+1)/(x^2+1)
=xarccotx+1/2ln|x^2+1|+C
=xarccotx+1/2ln(x^2+1)+C