Evaluate $\int_0^\infty \frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\mathrm dx$ How to evaluate 
$$\int_{0}^{\infty }\frac{\ln(2x+1)}{x(x+1)}\,\mathrm dx?$$
I tried
$$\frac{\ln(2x+1)}{x(x+1)}=\ln(2x+1)\left (\frac{1}{x}-\frac{1}{1+x}  \right)$$
but I don't know how to go on.

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\begin{align}
\int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x &
\,\,\,\stackrel{2x + 1\ \mapsto\ x}{=}\,\,\,
\int_{1}^{\infty}{\ln\pars{x} \over
\bracks{\pars{x - 1}/2}\bracks{\pars{x - 1}/2 + 1}}\,{\dd x \over 2} =
2\int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
-2\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 2}\int_{0}^{1}{x^{-1/2}\,\ln\pars{x} \over 1 - x}\,\dd x
\\[5mm] & =
\left.{1 \over 2}\,\totald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x\,
\right\vert_{\ \mu\ =\ -1/2} =
\left.{1 \over 2}\,\totald{\bracks{\Psi\pars{\mu + 1} + \gamma}}{\mu}
\right\vert_{\ \mu\ =\ -1/2}\label{1}\tag{1}
\end{align}

where $\ds{\Psi}$ is the $Digamma\ Function$ and $\gamma$ is the
  $Euler\mbox{-}Mascheroni\ Constant$: See $\mathbf{6.3.22}$ in A&S Table.

From expression \eqref{1}:
$$
\int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x =
{1 \over 2}\,\Psi\, '\pars{1 \over 2} = \bbx{\ds{\pi^{2} \over 4}}
$$

See $\mathbf{6.4.4}$ in
  A&S Table.

