Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$

$\bf{My\; try::}$ We can write  $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as 
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$
So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$
So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$ 
Now How can I solve it, Help me
Thanks 

$$\begin{eqnarray*}\sum_{k=0}^{+\infty}\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right)&=&\int_{0}^{1}\frac{1-x^3}{1-x^5}\,dx\\&=&\int_{0}^{1}\frac{1+\frac{1}{\sqrt{5}}}{2+(1-\sqrt{5})x+2x^2}\,dx+\int_{0}^{1}\frac{1-\frac{1}{\sqrt{5}}}{2+(1+\sqrt{5})x+2x^2}\,dx\end{eqnarray*} $$
where the last identity comes from the residue theorem.
By using the reflection formula for the $\psi$ function, we have that the original series equals:
$$ \frac{1}{5}\left(\psi\left(\frac{4}{5}\right)-\psi\left(\frac{1}{5}\right)\right) =\frac{\pi}{5}\cot\frac{\pi}{5}.$$
