In addition to the nice set of references by Raymond Manzoni, here is my proof of the identity. Frankly, I have not seen these references yet, thus I am not sure if this already appears in one of them.
Here I refer to the following identity
$$ \binom{\alpha}{\omega} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\omega \theta} \; d\theta, \ \cdots \ (1) $$
whose proof can be found in my blog post.
Now let $x$ be a real number such that $|x| < \frac{\pi}{2}$. Then simple calculation shows that
$$ \log\left(1+e^{2ix}\right) = \log(2\cos x) + ix \quad \Longleftrightarrow \quad \Im \left( \frac{-x}{\log\left(1+e^{2ix}\right)} \right) = \frac{x^2}{x^2 + \log^2(2\cos x)},$$
hence we have
$$ \begin{align*}I
&:= \int_{0}^{\frac{\pi}{2}} \frac{x^2}{x^2 + \log^2(2\cos x)} \; dx
= -\int_{0}^{\frac{\pi}{2}} \Im \left( \frac{x}{\log\left(1+e^{2ix}\right)} \right) \; dx \\
&= -\frac{1}{8}\int_{-\pi}^{\pi} \Im \left( \frac{\theta}{\log\left(1+e^{i\theta}\right)} \right) \; d\theta
= \frac{1}{8}\Re \left( \int_{-\pi}^{\pi} \frac{i\theta}{\log\left(1+e^{i\theta}\right)} \; d\theta \right).
\end{align*}$$
Differentiating both sides of $(1)$ with respect to $\omega$ and plugging $\omega = 1$, we have
$$ \frac{1}{2\pi} \int_{-\pi}^{\pi} (-i\theta) \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\theta} \; d\theta = \alpha \left(\psi_0(\alpha) - \psi_0(2)\right). $$
Now integrating both sides with respect to $\alpha$ on $[0, 1]$,
$$ \begin{align*}
-\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{i\theta}{\log \left(1 + e^{i\theta}\right)} \; d\theta
&= \int_{0}^{1} \alpha \left(\psi_0(\alpha) - \psi_0(2)\right) \; d\alpha \\
&= \left[ \alpha \log \Gamma (\alpha) \right]_{0}^{1} - \int_{0}^{1} \log \Gamma (\alpha) \; d\alpha - \frac{1}{2}\psi_0(2) \\
&= -\frac{1}{2}\left( 1 - \gamma + \log (2\pi) \right),
\end{align*}$$
where we have used the fact that
$$ \psi_0 (1+n) = -\gamma + H_n, \quad n \in \mathbb{N}$$
and
$$ \begin{align*}
\int_{0}^{1} \log \Gamma (\alpha) \; d\alpha
& = \frac{1}{2} \int_{0}^{1} \log \left[ \Gamma (\alpha) \Gamma (1-\alpha) \right] \; d\alpha \\
&= \frac{1}{2} \int_{0}^{1} \log \left( \frac{\pi}{\sin \pi \alpha} \right) \; d\alpha \\
&= \frac{1}{2} \left( \log \pi - \int_{0}^{1} \log \sin \pi \alpha \; d\alpha \right) \\
&= \frac{1}{2} \log (2\pi).
\end{align*} $$
Therefore we have the desired result.
Best Answer
Substituting $t=\ln x$ gives us that the integral $I$ satisfies $$I=\int_0^1\frac{e^tt}{(t+1)^2}dt.$$ Now we integrate by parts with $u=e^tt$ and $dv=\frac{1}{(t+1)^2}dt$. In particular, we get that $du=e^tt+e^t$ and $v=-\frac{1}{t+1}$. This implies that\begin{align*}I&=\int_0^1\frac{e^t(t+1)}{t+1}dt-\frac{e^tt}{t+1}\Bigg|_0^1\\&=-1+e-\frac e2=\boxed{\frac e2-1.}\end{align*}