TITLE: Integrate $\frac{1}{(1+x^2)(1+x^c)}$ from $0$ to $\infty$ for any $c$.
QUESTION [9 upvotes]: The question is to evaluate 
$$
\int_0^\infty \frac{dx}{(1+x^2)(1+x^c)}
$$
for arbitrary $c\geq0$. Here are my attempts:
(The methods behave somewhat differently for $c=0$ but that case is trivial so I will assume $c>0$)
Integrate by parts:
\begin{align*}
\int_0^\infty \frac{dx}{(1+x^2)(1+x^c)}&=\arctan(x)\frac{1}{1+x^c}\Big|_0^\infty+\int_0^\infty\arctan(x)\frac{cx^{c-1}}{(1+x^c)^2}dx\\
&=0+\int_0^\infty\arctan(x)\frac{cx^{c-1}}{(1+x^c)^2}dx\\
&=\int_0^\infty\frac{\arctan(x^{1/c})}{(1+x)^2}dx.
\end{align*}
This looks just as bad, if not worse, as before...
Partial fractions: good for small values of $c$, but seems impossible as $c$ gets large. 
Along the same line as partial fractions, but I tried the Ostragodski method, which only helped simplify things for $c=2$. 
Try and show it doesn't depend on $c$:
For $c=1$, $c=2$, and $c=3$, the integral evaluates to $\pi/4$. My gut tells me this isn't a coincidence, so define 
$$
f(c)=\int_0^\infty\frac{dx}{(1+x^2)(1+x^c)}
$$
so that 
$$
f'(c)=\int_0^\infty\frac{-x^c\log(x)}{(1+x^2)(1+x^c)^2}dx.
$$
It would be great if this evaluated to $0$ for arbitrary $c$, but I don't see a nice way to show that it does.
Any hints would be greatly appreciated. Thanks in advance.

REPLY [5 votes]: $$I(0)=\frac{\pi}{4}$$
$$\frac{d I}{dc}=-\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx$$
$$\frac{d I}{dc}\stackrel{x\rightarrow 1/x}{=}+\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx =-\frac{d I}{dc}$$
$$\frac{d I}{dc}=0 \Rightarrow I(c)=const=\frac{\pi}{4}$$