Answer
The series converges for all $x$ in the interval $[-1,1)$.
Work Step by Step
We apply the ratio test and by using L'Hopital rule, we have
$$
\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{3n+5}/\ln(n+1) }{ x^{3n+2}/\ln n}|\\
=|x^3|\lim _{n \rightarrow \infty} \frac{\ln n}{\ln(n+1)}=|x^3|\lim _{n \rightarrow \infty} \frac{n+1}{n}=|x^3|.
$$
Hence, the series $\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n$ converges if and only if $|x^3|\lt1$. That is, the interval of convergence is $(-1,1)$.
Now, we check the end points:
At $x=-1$, then
$\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n=\Sigma_{n=2}^{\infty}(-1)^{3n+2}/\ln n$, converges (Alternating Series Test).
At $x=1$, then
$\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n=\Sigma_{n=2}^{\infty}1/\ln n$, diverges (by the comparison test with $\frac{1}{\ln n}\geq \frac{1}{n}$).
Hence, the series converges for all $x$ in the interval $[-1,1)$.
