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.

Hi welcome to MSE:As a hint for telescopic series$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}
\\=\sum_{n=0}^{\infty}\frac{1}{(3n+1)(n+1)}
\\=\sum_{n=0}^{\infty}\frac{3}{(3n+1)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+2)}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+3)-1}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)}-3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$$
