Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$ 
Prove the following integral
  $$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$

This integral result was calculated using Mathematica and 
I like this integral. But I can't solve it. 
My idea:
Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$
so
$$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$
then I can't proceed. Can you help me? Thank you.

Let $f : [-1,1] \to \mathbb{R}$ be any continuous even function on $[-1,1]$. 
Consider following integral
$$I_f \stackrel{def}{=} 
\int_0^{\pi/2} f(\sin(\tan x)) dx 
= \frac12 \int_{-\pi/2}^{\pi/2} f(\sin(\tan x)) dx
= \frac12 \int_{-\infty}^{\infty} f(\sin y)\frac{dy}{1+y^2}
$$
where $y = \tan x$. Since $f(\cdot)$ is even, $f(\sin y)$ is periodic in $y$ with period $\pi$. We can rewrite $I_f$ as
$$
I_f = \frac12 \int_{-\pi/2}^{\pi/2} f(\sin y)g(y) dy
$$
where 
$$g(y) 
= \sum_{n=-\infty}^\infty \frac{1}{1+(y+n\pi)^2}
= \frac{1}{2i}\sum_{n=-\infty}^\infty \left(\frac{1}{n\pi + y - i} - \frac{1}{n\pi+y + i}\right)$$
Let $z = \tan y$ and $T = -i\tan i = \tanh 1 = \frac{e^2-1}{e^2+1}$. Recall following expansion of $\cot y$,
$$\cot y = \frac{1}{y} + \sum_{n \in \mathbb{Z} \setminus \{0\}}\left( \frac{1}{y+n\pi} - \frac{1}{n\pi}\right)$$
We find
$$g(y) = \frac{1}{2i}(\cot(y-i) - \cot(y+i))
= \frac{1}{2i} \left(\frac{1 + izT}{z - iT} - \frac{1 - izT}{z + iT}\right)
= \frac{(z^2+1)T}{z^2+T^2}
$$
As a result, we can get rid of the explicit trigonometric dependence in $I_f$:
$$
I_f = \frac12 \int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{(z^2+1)T}{z^2+T^2} \frac{dz}{1+z^2}
= \frac{T}{2}\int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{dz}{z^2+T^2}
$$
For any $a > 0$, let $b = \sqrt{a^2+1}$ and apply above formula to
$f_{a}(z) \stackrel{def}{=} \frac{1}{a^2+z^2}$, we have
$$\begin{align}
I_{f_a}
&= \frac{T}{2}\int_{-\infty}^\infty \frac{z^2+1}{a^2+b^2z^2}\frac{dz}{z^2+T^2}\\
&= \frac{T}{2(a^2-b^2T^2)}\int_{-\infty}^\infty \left(\frac{1-T^2}{z^2+T^2} - \frac{1}{a^2+b^2z^2}\right) dz\\
&= \frac{T}{2(a^2-b^2T^2)}\left((1-T^2)\frac{\pi}{T} - \frac{\pi}{ab}\right)
 = \frac{\pi}{2ab}\frac{b + aT}{a + bT}
\end{align}
$$
When $a = 1$, $b$ becomes $\sqrt{2}$ and $I_{f_1}$ reduces to the integral $I$ we want to compute, i.e.
$$I = I_{f_1} = \frac{\pi}{2\sqrt{2}}\frac{\sqrt{2}+\frac{e^2-1}{e^2+1}}{1+\sqrt{2}\frac{e^2-1}{e^2+1}}
= \frac{\pi}{2\sqrt{2}}\frac{(\sqrt{2}+1)e^2 + (\sqrt{2}-1)}{(\sqrt{2}+1)e^2 - (\sqrt{2}-1)}
= \frac{\pi}{2\sqrt{2}}\frac{e^2 + (\sqrt{2}-1)^2}{e^2 - (\sqrt{2}-1)^2}
$$
