Answer
The radius of convergence is $1$.
$\int \frac{t}{1-t^{8}}dt=c+\Sigma_{n=0}^{\infty}\dfrac{{t}^{8n+2}}{{t}^{8n+2}}$
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,
$\frac{t}{1-t^{8}}=\frac{a}{1-r}$
Therefore,
$f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(t)(t^{8})=\Sigma_{n=0}^{\infty}{t}^{8n+1}$
This is the power series representation of $f(t)$.
We know that the power series converges when $r=|t^{8}|\lt 1$
The interval of convergence is: $(-1,1)$
The radius of convergence is: $1$
$\int \frac{t}{1-t^{8}}dt=\int[ \Sigma_{n=0}^{\infty}{t}^{8n+1}]dt =c+\Sigma_{n=0}^{\infty}\dfrac{{t}^{8n+2}}{{t}^{8n+2}}$
