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.
 A: Alternatively we can use Feynman's Trick if we let, $$I(a)=\int_0^{\infty} \frac{\ln(ax+1)}{x(x+1)}\mathrm dx,$$ then if we take the derivative of $I(a)$ and expand using partial fractions we have  $$I'(a)=\int_0^{\infty} \frac{1}{(ax+1)(x+1)}\mathrm dx =\int_0^{\infty} \frac{a}{(a-1)(ax+1)} - \frac{1}{(a-1)(x+1)}\mathrm dx.$$  Integrating this gives, $$I'(a)=\bigg(\frac{1}{a-1}\ln|ax+1|-\frac{1}{a-1}\ln|x+1| \bigg)|_0^{\infty}=\frac{1}{a-1}\ln|\frac{ax+1}{x+1}| |_0^{\infty}.$$  Noticing the lower bound is zero we're left with an easy limit$$I'(a)=\lim_{x\to\infty}\frac{1}{a-1}\ln|\frac{ax+1}{x+1}|=\frac{\ln(a)}{a-1}.$$  Now if we integrate we can get $I(a)$, this is easy because our integral closely resembles the definition of the dilogarithm, so we have $$I(a)=\int \frac{\ln(a)}{a-1} da=-Li_2(1-a)+C.$$  Now if we notice $I(0)=0$ and note that $Li_2(1)=\frac{\pi^2}{6}$ we see that $C=\frac{\pi^2}{6}.$  So finally if we note that for your problem $a=2$ have that, $$I(2)=-Li_2(-1)+\frac{\pi^2}{6}=\frac{\pi^2}{12}+\frac{\pi^2}{6}=\frac{\pi^2}{4}.$$
This is admittedly not as nice as the approach using the geometric series and makes heavy use of the dilogarithm and it's particular values but I think it's kinda fun.
A: do the substitution $2x+1\rightarrow x$ and the let $x\rightarrow x^{-1}$ , hence
$$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\int_{1}^{\infty }\frac{\ln x}{x^{2}-1}\, \mathrm{d}x=-2\int_{0}^{1}\frac{\ln x}{1-x^{2}}\, \mathrm{d}x$$
then use the geometric series
$$\frac{1}{1-x^{2}}=\sum_{n=0}^{\infty }x^{2n}$$
we get
$$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\sum_{n=0}^{\infty }\frac{1}{\left ( 2n+1 \right )^{2}}$$
and the answer will follow.
A: Too long for a comment.
Making the problem more general, it is possible to compute $$I=\int \frac{\log (a x+b)}{(x+c) (x+d)}\,\mathrm dx$$ Using one integration by parts $$u=\log (a x+b)\implies u'=\frac{a}{a x+b}\,\mathrm dx$$ $$v'=\frac{\mathrm dx}{(c+x) (d+x)}\implies v=\frac{\log (x+d)-\log (x+c)}{c-d}$$ So, if $c\neq d$,$$(c-d)I=\log (a x+b) \log \left(\frac{(x+d) (a c-b)}{(x+c) (a
   d-b)}\right)-\text{Li}_2\left(\frac{a x+b}{b-a
   c}\right)+\text{Li}_2\left(\frac{a x+b}{b-a d}\right) $$ and, if $c=d$, $$I=\frac{a (x+c) \log (x+c)-(a x+b) \log (a x+b)}{(x+c) (b-a c)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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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.

A: This might be a late response, but consider the double integral
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dy \ dx.$$
We will evaluate $I$ in two ways. Integrating with respect to $y$ first, we see that
$$\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \ dy= \lim_{b \rightarrow \infty}\frac{\arctan((2b+1)x)}{2x}-\frac{\arctan(x)}{2x}=\frac{\pi}{4x}-\frac{\arctan(x)}{2x}$$ and get
$$I=\int_{0}^{\infty} \left(\frac{\pi}{4x}-\frac{\arctan(x)}{2x} \right) \frac{x}{1+x^2} \ dx=\int_{0}^{\infty} \frac{\pi}{4(1+x^2)}-\frac{\arctan(x)}{2(1+x^2)} \ dx=\frac{\pi^2}{8}-\frac{\pi^2}{16}=\frac{\pi^2}{16}.$$ 
Now we reverse the order of integration as such:
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dx \ dy.$$
To clean things up a little bit, let $u=x^2, \ du = 2 x \ dx.$ So 
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+u(2y+1)^2} \frac{u}{2(1+u)} \ du \ dy.$$
Integrating with respect to $u,$ we can use partial fractions to get that 
$$\frac{1}{1+u(2y+1)^2} \frac{1}{2(1+u)}=\frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)}.$$
We see 
$$\int_{0}^{\infty} \frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)} \ du $$
$$= \lim_{b \rightarrow \infty}\frac{\ln(b(2y+1)^2+1)}{2(2y+1)^2-2} - \frac{\ln(b+1)}{2(2y+1)^2-2}= \frac{\ln(2y+1)}{(2y+1)^2-1}=\frac{\ln(2y+1)}{4y(y+1)}.$$
Hence, 
$$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{4y(y+1)} \ dy=\frac{\pi^2}{16},$$
So $$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{y(y+1)} \ dy=\frac{\pi^2}{4}.$$
