How to integrate $x\ln(x+1)$?

Question

I am trying to compute $\int x\ln (x+1)\, dx$. I tried integrating by parts and ended up with: $$\int x\ln(x+1)\,dx = \frac{1}{2}x^2\ln(x+1) - \frac{1}{2}\int\frac{x^2}{x+1}\,dx$$ but I'm stuck here.

Answer 1

Write $x^2=x^2-1+1$ so that

$$\frac{x^2}{x+1}=\frac{x^2-1}{x+1}+\frac{1}{x+1}=(x-1)+\frac{1}{x+1} $$

Answer 2

Hints:View $x^2$ as $x^2-1+1$ in the second part.

Answer 3

$$\int \dfrac{x^2}{x+1}dx = \int \dfrac{x^2-1}{x+1}dx + \int \dfrac{dx}{x+1} = \int (x-1) dx + \int \dfrac{dx}{x+1} = \dfrac{x^2}2 - x + \ln(x+1)+ c$$