Aloha :)
Zum Integrieren hilft hier die partielle Integration:
$$\int\underbrace{(x+1)^2}_{=u'}\cdot\underbrace{\ln(x+1)}_{=v}\,dx=\underbrace{\frac{(x+1)^3}{3}}_{=u}\;\underbrace{\ln(x+1)}_{=v}-\int\underbrace{\frac{(x+1)^3}{3}}_{=u}\;\underbrace{\frac{1}{1+x}}_{=v'}\,dx$$$$\qquad=\frac{(x+1)^3}{3}\ln(x+1)-\int\frac{(x+1)^2}{3}dx=\frac{(x+1)^3}{3}\ln(x+1)-\frac{(x+1)^3}{9}+c$$$$\qquad\frac{1}{9}(x+1)^3\left(3\ln(x+1)-1\right)+c$$