1/1+x²的积分为什么不能用ln来算？

Question

如题所述

Answer 1

∫ x^2/(1+x^2) dx

=∫ [ 1- 1/(1+x^2) ]dx

=x-arctanx + C

Answer 2

∫1/(1+x²)=arctanx ＋C

Answer 3

∫ln(1+x^2)dx
分部积分法
∫ ln(1+x²) dx
=xln(1+x²)-∫ xd[ln(1+x²)]
=xln(1+x²)-∫ [x*2x/(1+x²)]dx
=xln(1+x²)-2∫ x²/(1+x²)dx
=xln(1+x²)-2∫ [1-1/(1+x²)]dx
=xln(1+x²)-2x+2arctanx+C本回答被网友采纳