Integrate $\frac{1}{(1+x^2)(1+x^c)}$ from $0$ to $\infty$ for any $c$. 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. 
 A: $$I(0)=\frac{\pi}{4}$$
$$\frac{d I}{dc}=-\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx$$
$$\frac{d I}{dc}\stackrel{x\rightarrow 1/x}{=}+\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx =-\frac{d I}{dc}$$
$$\frac{d I}{dc}=0 \Rightarrow I(c)=const=\frac{\pi}{4}$$
A: $$
I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^c)}
={{\pi} \over 4}$$
For any $c \gt 0$
Substitute 
$$x=\tan(u)$$
Cancel the $\sec(x)^2$
We'll get
$$\int_0^{\pi/2} {{1} \over {1+\tan^c(u)}} \ du$$
Symmetry can evaluate this integral
$$tan(\pi/2-u)=cot(u)$$
Change variables to $v=\pi/2-u$
We'll get $$I=\int_0^{\pi/2} {{\tan^c(u)} \over {tan^c(u)+1}} \ du$$
Add this to the first change of variables. Note the terms cancel and we get.
$$2 \cdot I=\int_0^{\pi/2} du$$
Which is $2 \cdot I=\pi/2$, which is the integral times two. Thus,
$$
\int_0^\infty \frac{dx}{(1+x^2)(1+x^c)}
={{\pi} \over 4}$$
For any $c$ actually
This trick can is further discussed here in the second answer.
(Posted on mobile)
A: For $c\ge0$,
$$\begin{align}
\mathcal{I}{\left(c\right)}
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}+\int_{1}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}+\int_{0}^{1}\frac{y^{c}\,\mathrm{d}y}{\left(1+y^{2}\right)\left(1+y^{c}\right)};~~~\small{\left[\frac{1}{x}=y\right]}\\
&=\int_{0}^{1}\frac{\left(1+x^{c}\right)\,\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{1+x^2}\\
&=\frac{\pi}{4}.\\
\end{align}$$
A: Sub $x=\tan{t}$.  Then the integral is
$$\int_0^{\pi/2} \frac{dt}{1+\tan^c{t}} $$
Then see this answer.
ADDENDUM
For those too tired to click the link:
Use the fact that
$$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$
i.e., 
$$\frac1{1+\tan^{c}{x}} = 1-\frac{\tan^{c}{x}}{1+\tan^{c}{x}} = 1-\frac1{1+\frac1{\tan^{c}{x}}} = 1-\frac1{1+\tan^{c}{\left (\frac{\pi}{2}-x\right)}}$$
Therefore, if the sought-after integral is $I$, then
$$I = \frac{\pi}{2}-I$$
and...
ADDENDUM II
It should be noted that the above result is valid for any real value of $c$, not just positive values.
