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

Because you seem to be looking for an approach without complex numbers; here is a not so neat, 'brute force' but real-valued approach.
Since $x^4 - x^3 + x^2 - x+ 1$ has no real roots, it has a factorization of the form:
$$\begin{align}
 x^4 \color{blue}{-1} x^3 \color{red}{+1} x^2 \color{green}{-1} x\color{purple}{+ 1}  & =\left( x^2+ax+b \right)\left( x^2+cx+d \right) \\
& = x^4 + \left( \color{blue}{a+c} \right)x^3 + \left(\color{red}{ac+b+d} \right)x^2 + \left( \color{green}{ad+bc}\right)x + \color{purple}{bd}
\end{align}$$
So you need to solve the system:
$$\left\{\begin{array}{rcr}
\color{blue}{a+c} & = & \color{blue}{-1} \\
\color{red}{ac+b+d} & = & \color{red}{1} \\
\color{green}{ad+bc} & = & \color{green}{-1} \\
\color{purple}{bd} & = & \color{purple}{1}
\end{array}\right.$$
Clearly $d=\tfrac{1}{b}$. If you suspect that a factorization exists (or if you'd simply try it) of the form $\left( x^2+ax+1 \right)\left( x^2+cx+1 \right)$, the system reduces to the far simpler:
$$\left\{\begin{array}{rcr}
a+c & = & -1 \\
ac+2 & = & 1 \\
\end{array}\right. \iff a = \frac{-1\pm\sqrt{5}}{2} \;,\; c = \frac{-1\mp\sqrt{5}}{2}$$
A: Beware: overkill. By Euler's Beta function and the reflection formulas for the $\Gamma$ function we have:
$$ \int_{0}^{+\infty}\frac{x^а\,dx}{1+x^б} = \frac{\pi}{б \sin\left(\frac{\pi(а+1)}{б}\right)} \tag{1}$$
as soon as $\text{Re}(а)>-1$ and $\text{Re}(б)>\text{Re}(а)+1$. Since $\sin\frac{2\pi}{5}=\sqrt{\frac{5+\sqrt{5}}{8}}$ it follows that:
$$ \int_{0}^{+\infty}\frac{x^2\,dx}{1+x^5} = \color{blue}{\frac{\pi}{5}\sqrt{2-\frac{2}{\sqrt{5}}}}\,.\tag{2}$$
A: 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.
A: Hint. Filling some details in @A.G. comment, notice that the roots of $x^5 = -1$ are $-1$ and the pairs of conjugate roots $z_1 = -e^{i\frac{2\pi}{5}}$ and $\bar{z}_1 = -e^{-i\frac{2\pi}{5}}$, and $z_2 = -e^{i\frac{4\pi}{5}}$ and $\bar{z}_2 = -e^{-i\frac{4\pi}{5}}$.
The first pair of conjugate roots is the solution of $x^2 + bx + c = 0$ where $c = z_1\bar{z}_1 = |z_1| = 1$ and $-b = z_1 + \bar{z}_1 = -2\cos(\frac{2\pi}{5})$.
Do the same to the pair $z_2$ and $\bar{z}_2$ to obtain the other second order polynomial of your decomposition.
A: 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*}
