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
Sum of geometric series with initial term $a$ and common ratio $r$ is
$S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
Here,
$\frac{tan^{-1}x}{x}=\frac{a}{1-r}$
Therefore,
$f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n}}{2n+1}$
This is the power series representation of $f(x)$.
we know that the power series converges when $r=|x^{2}|\lt 1$
Radius of convergence is $1$
$\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$
Radius of convergence is $1$
$\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$