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: $$\left\{\begin{array}{rcr} a+c & = & -1 \\ ac+2 & = & 1 \\ \end{array}\right. \iff a = \frac{-1\pm\sqrt{5}}{2} \;,\; c = \frac{-1\mp\sqrt{5}}{2}$$
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: $$ \int_{0}^{+\infty}\frac{x^2\,dx}{1+x^5} = \color{blue}{\frac{\pi}{5}\sqrt{2-\frac{2}{\sqrt{5}}}}\,.\tag{2}$$
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.
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.
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*}
$$ I=\frac{1}{5}\int_{\mathbb{R}_+}\frac{1}{x^{2/5}(1+x)}=\frac15B(\frac{1}{5},\frac45)=\frac{\Gamma(\frac15)\Gamma(\frac45)}{5} $$
where $B$ denotes Eulers Beta function. Using the reflection formula for the Gamma function on top of that yields the result
– tired Dec 16 '16 at 14:00