Is it possible to evaluate this integral using beta and gamma functions? There was an integral posted on Brilliant the other day, which is:
$$
\int_{0}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right)
\,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}
$$ 
I have seen the solution, but I was wondering if we could take a different approach and use gamma and beta functions instead. Would that be possible? 
Would it be possible to use this result? $$\int_0^{\infty} \frac{t^{x-1}}{(1+t)^{x+y}}dt= \beta(x,y)$$
Edited: After giving it some thought, there is no connection between the property I wrote above and the integral. However, after searching I have found this property: 
-$$\int_0^{\infty} \frac{t^{x-1}\ln(1+t)}{(1+t)^{x+y}}=\frac {\partial}{\partial y} \beta(x,y)$$
But I am still not very sure of how to apply it in order to solve that integral, or whether there are other properties we could perhaps use.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}}:\ {\Large ?}}$.

\begin{equation}
\mbox{Note that}\quad
\begin{array}{|l|}\hline\mbox{}\\
\ds{\quad\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} =\quad}
\\[3mm]
\ds{\quad\int_{0}^{\infty}
{\ln\pars{1 + x^{11}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x -
\int_{0}^{\infty}
{\ln\pars{1 + x^{3}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x\quad}
\\ \mbox{}\\ \hline
\end{array}
\label{1}\tag{1}
\end{equation}

With $\ds{\mu > 0}$:
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}
{\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x}}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{\infty}^{0}
{\ln\pars{1 + 1/x^{\mu}} \over \pars{1 + 1/x^{2}}\ln\pars{1/x}}
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
-\int_{0}^{\infty}
{\ln\pars{x^{\mu} + 1} - \mu\ln\pars{x}\over \pars{x^{2} + 1}\ln\pars{x}}\,\dd x
\\[5mm] \implies &\
\bbx{\int_{0}^{\infty}
{\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x =
{1 \over 2}\mu\int_{0}^{\infty}{\dd x \over 1 + x^{2}} = {1 \over 4}\,\mu\pi}
\label{2}\tag{2}
\end{align}

\eqref{1} and \eqref{2} lead to
$$
\bbx{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} =
{1 \over 4}\,11\pi - {1 \over 4}\,3\pi = {\large 2\pi}}
$$
A: We may consider that for any $\alpha>0$
$$ \frac{d}{d\alpha}\int_{0}^{+\infty}\frac{\log(1+x^\alpha)}{(1+x^2)\log x}\,dx =\int_{0}^{+\infty}\frac{x^{\alpha}}{(1+x^2)(1+x^{\alpha})}\,dx=\frac{\pi}{2}-\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^{\alpha})}$$
and with or without the Beta function it is well-known that $\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^{\alpha})}=\frac{\pi}{4}$ does not really depend on $\alpha$. It follows that $\int_{0}^{+\infty}\frac{\log(1+x^\alpha)}{(1+x^2)\log x}\,dx=\frac{\pi\alpha}{4}$ and
$$ \int_{0}^{+\infty}\frac{\log\left(\frac{1+x^{11}}{1+x^3}\right)}{(1+x^2)\log x}\,dx=\color{red}{2\pi}.$$

In any case,
$$ \int_{0}^{1}\frac{dx}{(1+x^2)(1+x^\alpha)}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/2}\frac{\cos^\alpha(\theta)}{\sin^\alpha(\theta)+\cos^\alpha(\theta)}d\theta\stackrel{\theta\mapsto\frac{\pi}{2}-\varphi}{=}\int_{0}^{\pi/2}\frac{\sin^\alpha(\varphi)}{\sin^\alpha(\varphi)+\cos^\alpha(\varphi)}d\varphi. $$
A: Let  $$ I=\int_{0}^{1}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right)
\,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}+\int_{1}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right)
\,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}$$
 $$\int_{1}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right)
\,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}=-\int_{0}^{1}\ln\left(\frac{1 + y^{11}}{1 + y^{3}}\right)
\,{\mathrm{d}y \over \left(1 + y^{2}\right)\ln\left(y\right)}+8\int_{0}^{1}\frac{dy}{1+y^2}$$
 Put $$x=\frac{1}{y}$$
 $$I=8\frac{\pi}{4}=2{\pi}$$
