Prove $\int_{0}^{1}{1\over 1+x^2}\ln\left({1+x\over 1-x}\right)=K$ Showing 

$$I=\int_{0}^{1}{1\over 1+x^2}\ln\left({1+x\over 1-x}\right)=K\tag1$$

$K=0.9159655...$ ;Catalan's constant.
Recall
$$\ln\left({1+x\over 1-x}\right)=2\sum_{n=0}^{\infty}{x^{2n+1}\over 2n+1}\tag2$$
Sub (2) into (1)$\rightarrow $(3)
$$I=2\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{1}{x^{2n+1}\over 1+x^2}dx\tag3$$
Can't go any further!

2nd approach

$$I=\int_{0}^{1}{1\over 1+x^2}\ln\left({1+x\over 1-x}\right)=K\tag1$$

Recall
$${1\over 1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag2$$
$$I=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^{2n}\ln\left({1+x\over 1-x}\right)dx\tag3$$
Applying integration by parts to (3)
Let $$J=\int_{0}^{1}x^{2n}\ln\left({1+x\over 1-x}\right)dx\tag4$$
$u=\ln\left({1+x\over 1-x}\right)\rightarrow du={2\over 1-x^2}dx$
$dv=x^{2n}\rightarrow v={x^{2n+1}\over 2n+1}$
$$J=\left.{x^{2n+1}\over 2n+1}\ln{1+x\over 1-x}\right|_{0}^{1}-{2\over 2n+1}\int_{0}^{1}{x^{2n+1}\over 1-x^2}dx\tag5$$
I don't is correct to integrate (4) in this manner, because when evaluating the limits it is not making sense.
Any help how to prove I?
 A: Enforcing the substitution $t=\frac{1-x}{1+x}$, we have $x=\frac{1-t}{1+t}$, $dx=-\frac{2}{(1+t)^2}\,dt$.  Therefore, 
$$\begin{align}
\int_0^1\frac{1}{1+x^2}\log\left(\frac{1+x}{1-x}\right)\,dx&=\int_1^0 \frac{(1+t)^2}{2(1+t^2)}\log(1/t)\left(-\frac{2}{(1+t)^2}\right)\,dt\\\\
&=-\int_0^1 \frac{1}{1+t^2}\log(t)\,dt\\\\
&=-\sum_{n=0}^\infty (-1)^n \int_0^1 t^{2n}\log(t)\,dt\\\\
&=\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)^2}\\\\
&=G
\end{align}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
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With the sub$\ldots$
$\ds{{1 + x \over 1 - x} \equiv t\ \iff\ x = {t - 1 \over t + 1}}$
\begin{align}
&\color{#f00}{%
\int_{0}^{1}{1 \over 1 + x^{2}}\ln\pars{{1 + x \over 1 - x}}\,\dd x} =
\int_{1}^{\infty}{\ln\pars{t} \over 1 + t^{2}}\,\dd t = \color{#f00}{K}
\end{align}
because it's a Catalan Constant integral representation.

Moreover, with $\ds{t \to {1 \over t}}$:

\begin{align}
&\color{#f00}{%
\int_{0}^{1}{1 \over 1 + x^{2}}\ln\pars{{1 + x \over 1 - x}}\,\dd x} =
-\int_{0}^{1}{\ln\pars{t} \over 1 + t^{2}}\,\dd t =
\Im\int_{0}^{1}{\ln\pars{t} \over \ic - t}\,\dd t =
\Im\int_{0}^{1}{\ln\pars{t} \over 1 - t/\ic}\,{\dd t \over \ic}
\\[3mm] = &\
\Im\int_{0}^{-\ic}{\ln\pars{t\ic} \over 1 - t}\,\dd t =
\Im\int_{0}^{-\ic}{\ln\pars{1 - t } \over t}\,\dd t =
-\Im\mathrm{Li}_{2}\pars{-\ic} =
-\Im\sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over n^{2}}
\\[3mm] = &\
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} = \color{#f00}{K}
\end{align}
A: Another approach. By setting $x=\tanh z$, then $z=\frac{u}{2}$, we get:
$$\begin{eqnarray*}I=\int_{0}^{1}\frac{2\text{arctanh}(x)}{1+x^2}\,dx&=&\int_{0}^{+\infty}\frac{2z}{\cosh^2 z+\sinh^2 z}\,dz\\&=&\int_{0}^{+\infty}\frac{u\,du}{e^u+e^{-u}}\\&=&\sum_{n\geq 0}(-1)^n\int_{0}^{+\infty}u e^{-(2n+1)u}\,du\\&=&\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}=\color{red}{K}\end{eqnarray*}$$
as wanted.
