<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] Here is a sketch of a way that does not use the digamma function.
If we start with your own favourite(?) change of variable,
$$
u=\frac{1}{x}-x,
$$
then, we end up with
$$
\int_0^{+\infty}\frac{2+u^2}{5+5u^2+u^4}\,du-\int_0^{+\infty}\frac{1}{\sqrt{4+u^2}(5+5u^2+u^4)}\,du
$$
In the first integral, it is easier (than in the original one) to do partial fraction decomposition. In the second integral, we let
$$
t=\frac{u}{\sqrt{4+u^2}},
$$
which will give us the integral
$$
\int_0^1\frac{1-t^2}{5+10t^2+t^4}\,dt,
$$
which also can be done using partial fraction decomposition (don't forget that it should come with a minus). I leave the funny calculations to you...
The final result (as already mentioned in several places before) turns out to be
$$
\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}.
$$
Probably there are smarter substitutions, leading to an even simpler expression to integrate.
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