<s>[INST] 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 
 [/INST] To expand the answer posted by @Arentino, we have
$$\begin{align}
S&=3\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+4)}\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k+4/5}\right)\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k}\right)+\frac15\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+4/5}\right)\\\\
&=\frac34+\frac15(\psi(9/5)-\psi(6/5)) \tag 1
\end{align}$$
Using the reflection and recurrence formulae of the digamma function in $(1)$ reveals that 
$$\begin{align}
S&=\frac34+\frac15(\psi(9/5)-\psi(6/5))\\\\
&=\frac34+\frac15\left(\psi(4/5)+\frac54\right)-\frac15\left(\psi(1/5)+5\right)\\\\
&=\frac34+\frac15\left(\psi(4/5)-\psi(1/5)\right)+\left(\frac14-1\right)\\\\
&=\frac15 \pi \cot(\pi/5)\\\\\
&=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}
\end{align}$$
as expected!!
</s>