Integral and limit: $\lim_{n \to \infty}\int_{-1}^{1}\frac{t{e}^{nt}}{{\left( {{e}^{2nt}}+1 \right)}^{2}}dt$ Show that, $$\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2}}\int_{-1}^{1}{\frac{t{{e}^{nt}}}{{{\left( {{e}^{2nt}}+1 \right)}^{2}}}dt=-\frac{\pi }{4}}$$
i reached this result after using  two steps of subsitution  to get the  following  are $x=nt$ then $dx=ndt$ 
$$\int_{-\infty}^{+\infty}\frac{xe^x}{(e^{2x}+1)^2}dx$$
i take $y=e^x$ then $dy=e^xdx$ hence i get 
$$  \int _0^ {\infty} \:\frac {ln\left (y\right)} {\left (y^2+1\right) ^2\:} dy$$
Now how shall i will continue  this problem?

Hint. Let's consider 
$$
\int _0^{\infty} \frac {\ln y} {y^2+a^2} \:dy,\qquad a>0,
$$ then, by the change of variable $y=ax$, $\ln y= \ln a +\ln x$, $dy=adx$, one has
$$
\int _0^{\infty} \frac {\ln y} {y^2+a^2} \:dy=\frac{\ln a}{a}\int _0^{\infty} \frac {1} {x^2+1} \:dx+\frac{1}{a}\int _0^{\infty} \frac {\ln x} {x^2+1} \:dx,
$$ the latter integral being equal to $0$ (make the change of variable $x \to 1/x$), this gives
$$
\int _0^{\infty} \frac {\ln y} {y^2+a^2} \:dy=\frac \pi2\frac{\ln a}{a}
$$ and, by differentiating with respect to $a$,
$$
\int _0^{\infty} \frac {\ln y} {(y^2+a^2)^2} \:dy=\frac \pi4\frac{\ln a}{a^3}-\frac{\pi }{4 a^3},\qquad a>0,
$$ by putting $a=1$ one obtains

$$
\int _0^{\infty} \frac {\ln y} {(y^2+1)^2} \:dy=-\frac{\pi}{4}.
$$

