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 $$ \int \frac{x^2}{1-x^5}\, \mathrm{d}x$$
 A: Well the way to do this is to just use partial fractions. The denominator can be factored into one linear and two quadratic factors.

Addendum: Not to steal anyone's answer but just to explain a bit more about what other people are saying. First like I said, the standard way to do these rational integrals is partial fractions. Second, the actual factoring in this case is not obvious so the easiest way to do it is what M. Strochyk said and end up with
$$x^5-1=(x-1)\left(x^2-2\cos{\left(\dfrac{2\pi}{5}\right)}x+1\right)\left(x^2-2\cos{\left(\dfrac{4\pi}{5}\right)}x+1\right)$$
and just for the sake of completion $\cos(2\pi/5)=\frac{-1+\sqrt{5}}{4}$ and $\cos(4\pi/5)=\frac{-1-\sqrt{5}}{4}$ and then to actually do the partial fraction decomposition and then integrating which is why the integrator gives you all those logs and arctangents. The logs come from the terms which have a degree difference of one between the numerator and the denominator and the arctangent terms are from the terms with a degree difference of two.
Third, to answer your actual question of finding $f(4)$ given that $f(1)=0$, you have asked us what is $f(4)$ given that
$$f(z)=\int_1^z \frac{x^2}{1-x^5}dx$$
BUT like Christian Blatter said, this is an improper integral because of the vertical asymptote at $x=1$ and this improper integral does NOT converge so your question has no answer. Even if you went through the pain of integrating this, $f(1)$ is undefined.
A: 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.
