Answer
The radius of convergence is $1$.
$\int \frac{t}{1+t^{3}}dt=c+\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{t}^{3n+2}}{3n+2}$
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,
$\frac{t}{1+t^{3}}=\frac{a}{1-r}$
Therefore,
$f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(t)(-t^{3})^{n}=\Sigma_{n=0}^{\infty}(-1)^{3}{t}^{3n+1}$
This is the power series representation of $f(t)$.
We know that the power series converges when $r=|t|\lt 1$
The radius of convergence is $1$.
$\int \frac{t}{1+t^{3}}dt=\Sigma_{n=0}^{\infty}(-1)^{3}{t}^{3n+1} =c+\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{t}^{3n+2}}{3n+2}$
