求∫In(1+x눀)dx

∫In(1+x²)dx=x*In(1+x²)+∫xd(In(1+x²))=x*In(1+x²)+∫2x²/(x²+1)dx=x*In(1+x²)+∫(2-2/(x²+1))dx
因为∫(1/(x²+1))dx=arctanx
所以 原式=x*In(1+x²)+2x-2arctanx+C

原式=xln(1+x²)-∫xdln(1+x²)
=xln(1+x²)-∫x*1/(1+x²)*2xdx

=xln(1+x²)-2∫x²/(1+x²)dx

=xln(1+x²)-2∫[1-1/(1+x²)]dx

=xln(1+x²)-2x+2arctanx+C

你好，很高兴回答的问题

∫ln(1+x^2)dx
=xln(1+x^2)-∫xdln(1+x^2)
=xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2) + ∫ [ -2 + 2/(1+x^2) ] dx
=xln(1+x^2) -2x + 2arctanx +C

∫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

利用分部积分就可以了