<s>[INST] 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?
 [/INST] 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.
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