stumbled here by accident ^^
Perhaps you were looking to use a particular method but this convergent series is a case of telescopic sum
\begin{eqnarray*}
S&\underset{def}=&\sum_{n\geq1}\frac1{9n^2+3n-2}\\
&=& \frac13\sum_{n\geq1}\Big(\frac1{3n-1}-\frac1{3(n+1)-1}\Big)\\
&=& \frac16,
\end{eqnarray*}
which is also the case of the series that appears in your integral, but this is not enough yet to exchange summation and integration.
If you really want to make sense of it, notice that the $N^{\text{th}}$ partial sum of the telescopic series integrated to any $y\in\mathbb R$ has value
\begin{eqnarray*}
\sum_{n=1}^N\int_0^y\big( x^{3n-1}-x^{3n+2}\big)dx&=&\int_0^y\sum_{n=1}^N\big( x^{3n-1}-x^{3n+2}\big)dx\\
&=&\frac{y^2}2-\frac{y^{3N+2}}{3N+2}.
\end{eqnarray*}
This now converges uniformly to $\frac{y^2}2$ when $N\rightarrow\infty$ on the compact set $|y|\leq1$, with boundary points now included. Uniform convergence implies continuity, letting you take the $y\rightarrow1$ limit.
It basically means that point-wise convergence of the telescopic series to a finite value at the boundary of the disc of convergence was only enough because we integrated it, in a way making it more regular. To see the effect of integration, notice that what made uniform convergence was the denominator $3N+2$ and that it appeared as a result of integration. "Telescopicity" was not enough, despite what one might have thought.
Hope it helps. Cheers, Raph.
