How to evaluate this definite integral: $\int_0^1 {(1-x^3)}/{(1-x^5)} \;\rm dx$ I need to compute:
$$\int_0^1 \frac{1-x^3}{1-x^5} dx$$
I tried integrating it by partial fractions but couldn't succeed. Is there any other way to integrate this?
 A: An alternative approach would be to use the geometric series
$$ \frac{1-x^3}{1-x^5} = \sum_{n=0}^{\infty} (x^{5n} - x^{5n+3}) $$
to evaluate the integral as
$$\int_0^1 \frac{1-x^3}{1-x^5} dx = \sum_{n=0}^{\infty} \Big(\frac{1}{5n+1} - \frac{1}{5n+4} \Big)$$
Using the digamma function this can now be evaluated as 
$$  \sum_{n=0}^{\infty} \Big(\frac{1}{5n+1} - \frac{1}{5n+4} \Big) = \frac{1}{5} (\psi(4/5) - \psi(1/5) ) $$
In conclusion one can note that the digamma function of a rational number in the interal $]0;1[$ can always be expressed as a sum of elementary functions (trigonometric functions and logarithms of trigonometric functions) evaluated at a rational number times $\pi$. 
A: Another method, using the Beta function. Let $t=x^5$, then
$$
\begin{align}
\int_0^1 \frac{1-x^3}{1-x^5} dx
&=\lim_{\delta\to0}\frac15\int_0^1(t^{-4/5}-t^{-1/5})(1-t)^{\delta-1}\,\mathrm{d}t\\
&=\lim_{\delta\to0}\frac15\left(\mathrm{B}(1/5,\delta)-\mathrm{B}(4/5,\delta)\right)\\
&=\lim_{\delta\to0}\frac15\left(\frac{\Gamma(1/5)\Gamma(\delta)}{\Gamma(1/5+\delta)}-\frac{\Gamma(4/5)\Gamma(\delta)}{\Gamma(4/5+\delta)}\right)\\
&=\lim_{\delta\to0}\frac15\left(\frac{\Gamma(1/5)}{\Gamma(1/5+\delta)}-\frac{\Gamma(4/5)}{\Gamma(4/5+\delta)}\right)\frac{\Gamma(1+\delta)}{\delta}\\
&=\frac15\left(\frac{\Gamma'(4/5)}{\Gamma(4/5)}-\frac{\Gamma'(1/5)}{\Gamma(1/5)}\right)\\
&=\frac15(\psi(4/5)-\psi(1/5))\\
&=\frac{\pi}{5}\cot\left(\frac{\pi}{5}\right)\\
&=\frac{\pi}{5}\sqrt{\frac{5+2\sqrt{5}}{5}}
\end{align}
$$
Using the identity $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$
A: Use partial fraction:
$$
\frac{1-x^3}{1-x^5} = \frac{\sqrt{5}-1}{\sqrt{5}(2x^2+(1+\sqrt{5})x+2)}+\frac{\sqrt{5}
+1}{\sqrt{5}(2x^2+(1-\sqrt{5})x+2)}
$$
and then integrate both summands with
$$
\int 1/(ax^2+bx+c)dx= 2\frac{\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}{\left(\sqrt{4ac-b^2}\right)} + \text{constant},
$$
which is done by completing the square and scaling/translating the denominator to the form $A^2y^2+C^2$. You'll get $\int 1 /(A^2y^2+C^2)dy = \frac{\tan^{-1}(Ay/C)}{AC}+\text{constant}$. 
Plugin your limits and you're done.
I leave the hard substitution work to you. Ship ahoi.
