为自己加油

∫ln(1+x²)dx

∫ln(1+x²)dx=xln(1+x²)-∫xdln(1+x²) =xln(1+x²)-∫x/(1+x²)2xdx =xln(1+x²)-2∫[(1+x²-1)/(1+x²)]dx =xln(1+x²)-2∫(1-1/(1+x²)dx =xln(1+x²)-2x+2arctanx+C.