<s>[INST] How to integrate $x\ln(x+1)$? 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.
 [/INST] $$\int \dfrac{x^2}{x+1}dx = \int \dfrac{x^2-1}{x+1}dx + \int \dfrac{dx}{x+1} = \int (x-1) dx + \int \dfrac{dx}{x+1} = \dfrac{x^2}2 - x + \ln(x+1)+ c$$
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