1,xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2)-2∫(1-1/(1+x^2))dx
=xln(1+x^2)-2(x-arctanx)
2,設(shè)t=√x,x=t^2,dx=2tdt
∫arctan√xdx
=∫2tarctantdt
=∫arctantd(t^2)
=t^2arctant-∫t^2/(1+t^2)dt
=t^2arctant-∫(1-1/(1+t^2)dt
=t^2arctant-t+arctant
=xarctan√x-√x+arctan√x