Question

Solve $\int xlog(1+x)dx$

Answer 1

$\int x\log |x+1|dx=\log |x+1|\int xdx-\int \frac{d\log 1+x}{dx}(\int xdx)dx=\frac{x^2}{2}\log |x+1|-\frac{1}{2}\int \frac{x^2}{1+x}dx=\frac{x^2}{2}\log |1+x|-\frac{1}{2}\int \frac{x^2-1+1}{x+1}dx=\frac{x^2}{2}\log |1+x|-\frac{1}{2}[\int \frac{(x+1)(x-1)}{x+1}dx+\int \frac{dx}{x+1}]==\frac{x^2}{2}\log |1+x|-\frac{1}{2}[\int (x-1)dx+\log |x+1|]=\frac{x^2}{2}\log |1+x|-\frac{1}{2}[\frac{x^2}{2}-x+\log |x+1|]=\frac{(x^2-1)}{2}\log |1+x|-\frac{x^2}{4}+\frac{x}{2}+C$