Answer
$\dfrac{\ln(2)}{2}-\dfrac{5}{18}+\ln (2)$
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 $dv=x^{2} dx \implies v=\dfrac{x^3}{3}$
$\displaystyle \int_{0}^1 x^2 \ln (x+1) \ dx= \ln (x+1) \dfrac{x^3}{3}-\int_0^1 (\dfrac{x^3}{3}) (\dfrac{1}{x+1})$
or, $=[\dfrac{x^2}{2} \ln (x+1) -\dfrac{x^3}{9}+\dfrac{x^2}{6}-\dfrac{x}{3}+\ln|x+1|]_0^1$
or, $=\dfrac{1^2}{2} \ln (1+1) -\dfrac{1^3}{9}+\dfrac{1^2}{6}-\dfrac{1}{3}+\ln|1+1|]-[\dfrac{0^2}{2} \ln (0+1) -\dfrac{0^3}{9}+\dfrac{0^2}{6}-\dfrac{0}{3}+\ln|0+1|]$
or, $=\dfrac{\ln(2)}{2}-\dfrac{1}{9}+\dfrac{1}{6}-\dfrac{1}{3}+\ln 2$
or, $=\dfrac{\ln(2)}{2}-\dfrac{5}{18}+\ln (2)$
