Evaluate $\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$

Question

This is a integration question from a previous calculus exam:

Evaluate $$\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$$

I rewrote it as $$\lim \limits_{b \to \infty} \int_0^b \left(\frac{x^{10}}{1+x^{14}} \right)^{2} \, dx,$$ where do I go on from there?

Answer 1

Hint:

$$\begin{align}\frac{1}{14}\int \frac{x^7 14x^{13}}{(1 + x^{14})^2}dx& \underbrace{=}_{\color{red}{\text{by parts}}}\frac{1}{14}\bigg( -x^7\frac{1}{1 + x^{14}} + 7\color{#05f}{\int \frac{x^6}{1+x^{14}}dx}\bigg) \\&= \frac{1}{14}\bigg(\arctan(x^7) -\frac{x^7}{1 + x^{14}}\bigg) + C\end{align}$$

For $$\color{#05f}{\int \frac{x^6}{1+x^{14}}dx = \frac{1}{7}\arctan (x^7) + C}$$ let $v = x^7$.

Answer 2

In general, $~\displaystyle\int_0^\infty\frac{x^{k-1}}{a^n+x^n}~dx~=~a^{k-n}~\frac\pi n~\csc\bigg(k~\frac\pi n\bigg).~$ Now all that's left to do is

differentiating both sides with regard to a, and giving appropriate values to the three

parameters. The initial identity can be proven by letting $x=at$ and $u=\dfrac1{1+t^n}~,~$

followed by recognizing the expression of the beta function in the new integral, and

applying Euler's reflection formula for the $\Gamma$ function to arrive at the desired result.