<s>[INST] How to integrate this integral 
Let $$f\left(x\right)= \mbox{ the antiderivative } \frac{x^2}{1-x^5}, $$
$f\left(1\right)=0$
Find $f\left(4\right)$

I know that:
$F\left(x\right) = I\left(x\right) + C $ where $I\left(x\right)$ is the antiderivative (integral) of $f\left(x\right)$
Thus at $x = 1\; F\left(1\right) = 0 = I\left(1\right) + C$, thus $C = -I\left(x\right)$
Thus $F\left(4\right) = f\left(4\right) - I\left(1\right)$
My problem, is how to find the integral of $$ \int \frac{x^2}{1-x^5}\, \mathrm{d}x$$
 [/INST] Equation $z^5-1=0$ has one real root $z_0=1$ and two pairs of complex-conjugated roots $z_k=e^{\tfrac{2k\pi i}{5}},\;\; k=1,\,\ldots\,,4.$ Then 
$$
z^5-1=(z-z_0)(z-z_1)(z-\bar{z}_1)(z-z_2)(z-\bar{z}_2)=\\
=(z-1)(z^2-2\Re{(z_1)} z+1)(z^2-2\Re{(z_2)} z+1)=\\
=(z-1)\left(z^2-2\cos{\left(\dfrac{2\pi}{5}\right)} z+1\right)\left(z^2-2\cos{\left(\dfrac{4\pi}{5}\right)} z+1\right).$$
It is a factorization suggested by Fixed Point.
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