Question

求∫In(1+x²)dx 用分部积分答案不对

Answer 1

∫In(1+x²)dx
=xln(1+x²)-2∫x²/(1+x²)dx
=xln(1+x²)-2∫(x²+1-1)/(1+x²)dx
=xln(1+x²)-2∫{1-1/(1+x²)}dx
=xln(1+x²)-2∫dx+∫2/(1+x²)}dx
=xln(1+x²)-2x+2arctanx+C