原式=xln(x²+1)-∫xdln(x²+1)
=xln(x²+1)-∫2x²/(x²+1)dx
=xln(x²+1)-2∫(x²+1-1)/(x²+1)dx
=xln(x²+1)-2∫[1-1/(x²+1)]dx
=xln(x²+1)-2x+2arctanx+C
求一道不定积分分部积分的题.∫ ln(X²+1)dx
原式=xln(x²+1)-∫xdln(x²+1)
=xln(x²+1)-∫2x²/(x²+1)dx
=xln(x²+1)-2∫(x²+1-1)/(x²+1)dx
=xln(x²+1)-2∫[1-1/(x²+1)]dx
=xln(x²+1)-2x+2arctanx+C