Integral $\int_{-\infty}^{+\infty} \frac { e^{t}( e^{2t} -1)} {t(1+e^{2t})^2 }dt$ 
How to prove that
  $$\int_{-\infty}^{+\infty}  \frac  { \pi e^{t}( e^{2t} -1)} {4t(1+e^{2t})^2 }d t=C$$
  where $C$ is Catalan's constant.

Here is one of my ideas. Let: $$F(a)=\int_{-\infty}^{+\infty}  \frac  { \pi e^{at}( e^{2t} -1)} {4t(1+e^{2t})^2 }d t$$ for $0<a\leq 1$, then $$\displaystyle F'(a)=\int_{-\infty}^{+\infty}  \frac  { \pi e^{at}( e^{2t} -1)} {4(1+e^{2t})^2 }d t$$
Now put  $x=e^t$ then $$\displaystyle F'(a)=\int_{0}^{+\infty}  \frac  { \pi x^{a-1}( x^{2} -1)} {4(1+x^{2})^2 }dx$$
put $y=x^2$, then $$\displaystyle F'(a)=\frac {\pi}8 \int_{0}^{+\infty}  \frac  {  y^{\frac a2}( y -1)} {(1+y)^2 }d y =?$$
 A: $$I=\int_{-\infty}^\infty \frac{e^t(e^{2t}-1)}{t(1+e^{2t})^2}dt\overset{e^t=x}=\int_0^\infty \frac{(x^2-1)}{(x^2+1)^2 \ln x}dx$$
Consider: $$I(a)=\int_0^\infty \frac{(x^a-1)}{(x^2+1)^2 \ln x}dx\Rightarrow I'(a)=\int_0^\infty \frac{x^a}{(x^2+1)^2}dx\overset{x^2=y}=\frac12 \int_0^\infty \frac{y^{a/2}y^{-1/2}}{(y+1)^2}dy$$
$$=\frac12B\left(\frac{a+1}{2},2-\frac{a+1}{2}\right)=\frac12 \Gamma\left(\frac{a+1}{2}\right)\Gamma\left(2-\frac{a+1}{2}\right)=\frac12 \left(\frac{1-a}{2}\right)\frac{\pi}{\sin\left(\frac{\pi(a+1)}{2}\right)}$$
But we have that $I(0)=0$, so integrating back we need to find:
$$I=\frac{\pi}{4}\int_0^2 \frac{1-a}{\sin\frac{\pi(a+1)}{2}}da=\frac1{\pi} \int_\frac{\pi}{2}^\frac{3\pi}{2} \frac{\pi-y}{\sin y}dy =\frac{2}{\pi} \int_{0}^\frac{\pi}{2}\frac{t}{\sin t}dt$$
$$\overset{IBP}=-\frac{2}{\pi}\int_0^\frac{\pi}{2}\ln\left(\tan\frac{t}{2}\right)dt\overset{\tan \frac{t}{2}=e^x}=-\frac{4}{\pi}\int_0^1 \frac{\ln x}{1+x^2}dx$$
$$=-\frac{4}{\pi}\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac{4}{\pi}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=\frac{4 G}{\pi}$$
