#### Answer

The radius of convergence is $1$
$\int \frac{tan^{-1}x}{x}dx=c+\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n+1}}{(2n+1)^{2}}$

#### 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,
$\frac{tan^{-1}x}{x}=\frac{a}{1-r}$
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 $r=|x^{2}|\lt 1$
The radius of convergence is $1$
$\int \frac{tan^{-1}x}{x}dx=c+\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n+1}}{(2n+1)^{2}}$