How to evaluate the sum of $\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$ I hava an infinite sum
$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$
I factored the denominator
$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$
Then I separated the fraction
$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)}-\frac{1}{\left(n+1\right)}$$
Then I set 1 (the numerator) to be equal to x to some power which I don't know if I can do
$$\frac{3}{2}\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$
Then I set the integral which would satisfy the previous terms
$$\frac{3}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{3n}dx$$
and
$$-\frac{1}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{n}dx$$
Then I changed the order of summation and integration and I got
$$\frac{3}{2}\int_{0}^{1}\frac{1}{1-x^{3}}dx$$ and $$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
The first integral can be factored to
$$\frac{3}{2}\int_{0}^{1}\frac{1}{\left(1-x\right)\left(1+x+x^{2}\right)}dx$$
Then separated
$$\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}+\frac{x+2}{1+x+x^{2}}dx$$
The first one will cancel out with
$$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
And I'm left with
$$\frac{1}{2}\int_{0}^{1}\frac{x+2}{1+x+x^{2}}dx$$
which is
$$\frac{\sqrt{3}\pi}{12}+\frac{\ln\left(3\right)}{4}$$
But the correct answer is
$$\frac{\sqrt{3}\pi}{12}+\frac{3\ln\left(3\right)}{4}$$
So I would like to ask if this approach is invalid or if I'm just missing something.

$$ \begin{align}\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{3}{3n+1}-\frac{1}{n+1} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \left(3\int_{0}^{1}x^{3n} \, \mathrm dx- \color{red}{3\int_{0}^{1} x^{3n+2} \, \mathrm dx}\right) \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} \left(x^{3n}-x^{3n+2} \right) \, \mathrm dx \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} x^{3n}\left(1-x^{2}\right) \, \mathrm dx \end{align}$$
Since $x^{3n}(1-x^{2})$ is nonnegative for all $n$ and all $x$, Tonelli's theorem allows us to interchange the order of integration and summation.
Therefore, we have $$ \begin{align} \sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{3}{2} \int_{0}^{1} (1-x^{2}) \sum_{n=0}^{\infty} x^{3n}\, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{1-x^{2}}{1-x^{3}} \, \mathrm dx  \\ &= \frac{3}{2} \int_{0}^{1} \frac{x+1}{x^{2}+x+1} \, \mathrm dx \\ &= \frac{3}{4} \int_{0}^{1} \left( \frac{2x+1}{x^{2}+x+1} +\frac{1}{x^{2}+x+1} \right)\, \mathrm dx \\ &=  \frac{3}{4} \, \ln(3) + \frac{3}{4} \int_{0}^{1} \frac{\mathrm dx}{\left(x+\frac{1}{2} \right)^{2} + \frac{3}{4}} \, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{2\sqrt{3}} \, \arctan \left(\frac{2}{\sqrt{3}} \left( x+\frac{1}{2}\right) \right) \Bigg|_{0}^{1} \\ & = \frac{3}{4} \, \ln(3) + \frac{3}{2 \sqrt{3}} \left(\frac{\pi}{3}- \frac{\pi}{6} \right) \\ &=\frac{3}{4} \, \ln(3) + \frac{\sqrt{3} \pi}{12}. \end{align}$$
