∫xlog|x+1|dx=log|x+1|∫xdx−∫dlog1+xdx(∫xdx)dx=x22log|x+1|−12∫x21+xdx=x22log|1+x|−12∫x2−1+1x+1dx=x22log|1+x|−12[∫(x+1)(x−1)x+1dx+∫dxx+1]==x22log|1+x|−12[∫(x−1)dx+log|x+1|]=x22log|1+x|−12[x22−x+log|x+1|]=(x2−1)2log|1+x|−x24+x2+C