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. 

Substituting $x\mapsto\frac1x$, we get
$$
\begin{align}
\int_1^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{x\,\mathrm{d}x}{1+x^5}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{\left(x^2+x\right)\mathrm{d}x}{1+x^5}\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\left(x^{5k+1}+x^{5k+2}\right)\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\left(\frac1{5k+2}+\frac1{5k+3}\right)\\
&=\sum_{k=1}^\infty\left(\frac1{10k-8}+\frac1{10k-7}-\frac1{10k-3}-\frac1{10k-2}\right)\\
&=\frac1{10}\sum_{k=1}^\infty\left(\frac1{k-\frac8{10}}+\frac1{k-\frac7{10}}-\frac1{k-\frac3{10}}-\frac1{k-\frac2{10}}\right)\\[3pt]
&=\frac1{10}\left(H_{-\frac3{10}}+H_{-\frac2{10}}-H_{-\frac8{10}}-H_{-\frac7{10}}\right)\\[9pt]
&=\frac\pi{10}\left(\cot\left(\frac{3\pi}{10}\right)+\cot\left(\frac{2\pi}{10}\right)\right)\\[6pt]
&=\frac\pi5\sqrt{2-\frac2{\sqrt5}}
\end{align}
$$
Using $(5)$ from this answer, which is probably out of scope.
