Giải thích các bước giải:
Ta có :
$\int x\ln(x+1)dx$
$=\int \left(u-1\right)\ln \left(u\right)du, u=x+1$
$=\int\ln \left(u\right)d(\dfrac{u^2}2-u)$
$=\ln \left(u\right)\left(\frac{u^2}{2}-u\right)-\int \frac{1}{2}\left(u-2\right)du$
$=\ln \left(u\right)\left(\frac{u^2}{2}-u\right)-\left(\frac{u^2}{4}-u\right)$
$=\ln \left(x+1\right)\left(\frac{\left(x+1\right)^2}{2}-\left(x+1\right)\right)-\left(\frac{\left(x+1\right)^2}{4}-\left(x+1\right)\right)+C$
$\to\int^{e-1}_0 x\ln(x+1)dx=\dfrac{e^2}{4}-\dfrac{3}{4}$
