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. 


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:
$$\left\{\begin{array}{rcr}
a+c & = & -1 \\
ac+2 & = & 1 \\
\end{array}\right. \iff a = \frac{-1\pm\sqrt{5}}{2} \;,\; c = \frac{-1\mp\sqrt{5}}{2}$$
