Answer
$\dfrac{1}{4}-\dfrac{\ln (2)}{2}$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln (x+1)$ and $dv=x \ dx \implies v=\dfrac{x^2}{2}$
$\displaystyle \int_{0}^1 x \ln (x+1) \ dx= \ln (x+1) (\dfrac{x^2}{2}) -\int_0^1 (\dfrac{x^2}{2}) (\dfrac{1}{x+1}) \ dx$
or, $= (\dfrac{x^2}{2}) \ln (x+1)- \dfrac{1}{2} \int_0^1 \dfrac{x^2}{x+1} \ dx$
or, $=[(\dfrac{x^2}{2}) \ln (x+1) - \dfrac{x^2}{4}+\dfrac{x}{2}-\ln |x+1|]_0^1$
or, $=[(\dfrac{1^2}{2}) \ln (1+1) - \dfrac{1^2}{4}+\dfrac{1}{2}-\ln |1+1|]-[(\dfrac{0^2}{2}) \ln (0+1) - \dfrac{0^2}{4}+\dfrac{0}{2}-\ln |0+1|]$
or, $=\dfrac{1}{4}-\dfrac{\ln (2)}{2}$
