I want to find the value of $$\int_0^{\pi /2}\dfrac{u^2\ln{(2\cos u)}}{(u^2+\ln^2{(2\cos u)})^2}du.$$
Let $v=\frac{\pi}{2}-u$, then $$\int_0^{\pi /2}\dfrac{u^2\ln{(2\cos u)}}{(u^2+\ln^2{(2\cos u)})^2}du=\int_0^{\pi /2}\dfrac{(\pi /2-v)^2\ln{(2\sin v)}}{((\pi /2-v)^2+\ln^2{(2\sin v)})^2}dv.$$
I don't know how to go from here. Thank you.
One may find the following result.
Proposition. $$ \bbox[15px,border:1px solid #FF6600]{\int_0^{\pi/2}\!\! \frac{u^2\log(2\cos u)}{(u^2 + \log^2(2\cos u))^2}\: \mathrm{d}u =\frac{7\pi}{192} + \frac{\pi}{96}\ln (2 \pi) - \frac{\zeta'(2)}{16\pi}}\tag1 $$
where $\zeta(\cdot)$ is the Riemann zeta function. A numerical value of the preceding expression is $$ \color{#FF6600}{0.193333548399328789605977464282949669744336\cdots}.$$
Here is an approach.
Let $u$ be any real number such that $0<u< \dfrac{\pi}2$. Then from $$\log\left(1+e^{2iu}\right) = \log(2\cos u) + iu \tag2$$ we have $$ \begin{align} \frac{-2u\log(2\cos u)}{(u^2 + \log^2(2\cos u))^2}&=\Im \left(\frac1{\log^2\left(1+e^{2iu}\right)}-e^{-4iu}-e^{-2iu} \right)\\\\ &=\Im \left[\left(\frac1{\log^2\left(1-z\right)}-\frac1{z^2}+\frac1z \right)_{\large z=-e^{2iu}}\right]\tag3\\\\ &=\sum_{n=0}^\infty(-1)^n\frac{\alpha_n}{n!}\:\sin (2n u) \tag4 \end{align} $$ where we have set $$ \frac1{\log^2\left(1-z\right)}-\frac1{z^2}+\frac1z =\sum_{n=0}^\infty\frac{\alpha_n}{n!}z^n \qquad |z|\leq1,\,z \neq1. \tag5 $$ Observing that, as $n \to \infty$, $$ \frac{\alpha_n}{n!} \sim \mathcal{O}\left(\frac1{n \ln^3 n}\right),\tag6 $$ which may be obtained by noticing that $$ \frac1{\log^2\left(1-z\right)}-\frac1{z^2}+\frac1z=(1-z)\frac{d}{dz}\left(\frac1{\log\left(1-z\right)}+\frac1z \right),\tag7 $$ and observing that $$ \int_0^{\pi/2} u \sin ( 2 n u) \: \mathrm{d}u = \frac{\pi}4{\frac{(-1)}{n}\!}^{n-1} \qquad n = 1, 2, ...,\tag8 $$ then, multiplying $(4)$ by $u$, we are allowed, using $(6)$, to integrate $(4)$ termwise obtaining $$ \frac8{\pi}\int_0^{\pi/2}\!\! \frac{u^2\log(2\cos u)}{(u^2 + \log^2(2\cos u))^2}\: \mathrm{d}u =\frac12+\sum_{n=1}^\infty\frac{\alpha_n}{n!}\frac1n. \tag9 $$
One may observe, using $(5)$, noticing that $\alpha_0=\dfrac1{12}$ and performing a legitimate termwise integration, that $$ \int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{\mathrm{d}x}x=\sum_{n=1}^\infty\frac{\alpha_n}{n!}\frac1n.\tag{10} $$ Thus, from $(9)$ and $(10)$, we have $$ \frac8{\pi}\!\int_0^{\pi/2}\!\! \frac{u^2\log(2\cos u)}{(u^2 + \log^2(2\cos u))^2}\: \mathrm{d}u =\frac12+\!\!\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{\mathrm{d}x}x.\tag{11} $$
To deduce a closed form of the latter integral, one may introduce a parameter as followed. Let us consider $$ I(s):=\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{(1-x)^s}x\:\mathrm{d}x, \qquad s>0.\tag{12} $$ By differentiating twice, one gets rid of the $\log^2$ term in the denominator, then one has $$ I''(s)=-\frac{3}{2}-\frac{1}{s}+\frac{1}{6 s^3}+\left(2s+1\right)\psi'(s)+\left(\frac{s^2}{2}+\frac{s}{2}+\frac1{12}\right)\psi''(s)\tag{13}, $$ from which, integrating twice, using $I(s) \to 0$ as $s \to \infty$, we get the following new result.
Theorem. Let $s>0$. Then $$\begin{align} &\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{(1-x)^s}x\:\mathrm{d}x\\\\ &=\frac7{24}+\ln A+\frac1{12 s}+\left(\frac34+\frac12 \ln (2 \pi)\right)s-\frac{3 s^2}{4}\tag{14}\\\\ &-s \ln s+\frac1{12} \left(1+6 s+6 s^2\right) \psi(s)-\int_0^s\log \Gamma(t)\:dt. \end{align} $$
where $A$ is the Glaisher-Kinkelin constant.
Thus, by letting $s \to 0^+$ in $(14)$ and using $(11)$, taking into account that $$\ln A=\frac1{12}(\gamma+\log(2\pi))-\frac{\zeta'(2)}{2\pi ^2},\tag{15}$$ we obtain $(1)$, as announced.
