Computing $\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}$ 
How can I compute $$\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}\quad?$$ 

My attempt: Looking at the limits of the integration I see that we should induce some $\tan^{-1}(x)$ so if we put $\infty$ we would get something like $\frac{\pi}{2}$ . But I am not sure how to proceed . Partial fractions don't yield good integral.
 A: For any $a > 0$, consider the integral
$$\mathcal{I} = \int_0^\infty \frac{1}{1+x^a} \frac{dx}{1+x^2}$$
Change variable to $y = \frac1x$, we have
$$\mathcal{I} = \int_0^\infty \frac{1}{1+y^{-a}}\frac{1}{1+y^{-2}} \frac{dy}{y^2}
= \int_0^\infty \frac{y^a}{1+y^a}\frac{dy}{1+y^2}$$
Summing the two expression of $\mathcal{I}$ and divide by $2$, we get
$$\mathcal{I} = \frac12 \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{4}$$
independent of value of $a$.

Random Notes
About the question how do I arrive this answer, the key is symmetry.
When one is facing a complicated looking integral, the first thing one should do is looking for symmetry hidden in the integral.
For the integral at hand, the integrand has similar form and the range of integration is invariant under a change of variable from $x$ to $\frac1x$.
In this case, there are a few things one should check.


*

*Do different part of integral cancel each other?

*Can the original and transformed integral combined together to something simpler?

*Will changing variable to $y = x \pm \frac{\text{const}}{x}$ reduces the integrand to a more manageable form.


It turns out the second strategy works and the rest is mostly following your nose.
A: HINT:
Enforce the substitution $x\to 1/x$ and combine the results.
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution.

 Enforcing the substitution $x\to 1/x$ reveals $$\begin{align}I&=\int_0^\infty \frac{1}{(1+x^{2015})(1+x^2)}\,dx\\\\&=\int_0^\infty \frac{x^{2015}}{(1+x^{2015})(1+x^2)}\,dx\tag 1\end{align}$$Adding the right-hand and left-hand sides of $(1)$ and dividing by $2$ yields $$I=\frac12\int_0^\infty \frac{1}{1+x^2}\,dx=\pi/4$$

