Generalization to this integral $$ \int_0^\infty \frac{\ln(1 + x^a)x^s}{1+x^2} \ dx $$
Actually the problem was $ \displaystyle \int_0^\infty \frac{\ln(1 + x^a)}{(1+x^2)\ln(x)} \ dx $. 
But I guess the form of a Mellin Transform would be much better. Also, the above form would allow us to use Complex Analysis in a much easier way. 
I don't know how to do this problem. I have tried using Ramanujan Master Theorem but I got stuck in finding a definite series. I have guessed that there is an easier way to approach to this problem using Contour integration but I just don't know how. Please help!
 A: At least Maple agrees that
$$
\int_0^\infty \frac{\ln(1 + x^a)x^s}{1+x^2} \, dx
$$
is non-trivial.
Even with $s=0$.
Maple does
$$
\int_0^\infty \frac{\ln(1 + x^2)}{1+x^2} \ dx\quad\text{and}\quad
\int_0^\infty \frac{\ln(1 + x^4)}{1+x^2} \, dx
$$
OK, but
$$
\int_0^\infty \frac{\ln(1 + x^3)}{1+x^2} \, dx
$$
in terms of dilogarithms.  And the problem
$$
\int_0^\infty \frac{\ln(1 + x^5)}{1+x^2} \, dx
$$
is still running...
added
I stopped it.  Actually
$$
\int_0^\infty \frac{\ln(x-w)}{1+x^2} \, dx
$$
is evaluated in terms of dilogs, and and
$$
\int_0^\infty \frac{\ln(1 + x^5)}{1+x^2} \, dx
$$
is a sum of 5 of those.  For the full problem,
$$
\int_0^\infty \frac{\ln(x-w)x^s}{1+x^2} \, dx,
$$
Maple does it it terms of the Lerch Phi function.
$$
\int _{0}^{\infty }\!{\frac {\ln  \left( x-w \right) {x}^{s}}{{x}^{2}+
1}}{dx}= \\ \pi \, \left[ 2\,\ln  \left( -w \right) w\sin \left( 1/2
\,\pi \,s \right) +2\,w\arctan \left( {w}^{-1} \right) \cos \left( 1/2
\,\pi \,s \right) +\sin \left( 1/2\,\pi \,s \right) w\ln  \left( {w}^{
2}+1 \right) +\sin \left( 1/2\,\pi \,s \right) w\ln  \left( {w}^{-2}
 \right) - \left( -{w}^{-1} \right) ^{-s}\Phi \left( -{w}^{-
2},1,1/2-1/2\,s \right)  \right]\\  \left( \cos \left( 1/2\,\pi \,s
 \right)  \right) ^{-1} \left( \sin \left( 1/2\,\pi \,s \right) 
 \right) ^{-1}{w}^{-1} 4^{-1}
$$
