Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to. 

Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
  using only real-analytic techniques. 

My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand. 

Hint
$$x^4 - x^3 + x^2 - x+ 1=\left(x^2-\frac{\sqrt{5}+1}{2}x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2}x+1\right)$$
Now apply the partial fraction decomposition method.
Edit
\begin{align*}x^4 - x^3 + x^2 - x+ 1&=x^2\left(x^2-x+1-\frac{1}{x}+\frac{1}{x^2}\right)\\
& =x^2\left((x^2+\frac{1}{x^2})-(x+\frac{1}{x})+1\right)\\
&=x^2\left((x+\frac{1}{x})^2-(x+\frac{1}{x})-1\right)\\
&=x^2\left(x+\frac{1}{x}-\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{1-\sqrt{5}}{2}\right)\\
&=\left(x^2-\frac{\sqrt{5}+1}{2}x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2}x+1\right)
\end{align*}
