Answer
Radius of convergence is $1$
$\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$
Work Step by Step
The sum of a geometric series with initial term $a$ and common ratio $r$ is
$S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
Here,
$x^{2}ln(1+x)=\frac{a}{1-r}$
We know that the power series converges when $r=|x|\lt 1$
Hence, the radius of convergence is $1$
$\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$
