<s>[INST] How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$ 
Evaluate
  $$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$  

I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got  
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$  
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.  
Please Help!
 [/INST] Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. Now you post this question which I believe it's taken from one of proposed problems in Brilliant Integration Contest - Season 1 that I held on Brilliant.org. The original problem was proposed by Jatin Yadav as PROBLEM 7 but a day later he deleted this problem and changed to another problem after no-one can solve it included himself. According to him, it's taken from here, on Math S.E. You may want to take a look there.
I tried to solve this problem for hours but no success. Here is my attempt:
Set $x=\tan y$, we get
\begin{align}
I&=\int_0^1\arctan^2x\,\ln\left(\frac{1+x^2}{2x^2}\right)\,dx\\
&=-\int_0^{\pi/4} \frac{y^2\ln\left(2\sin^2y\right)}{\cos^2y}\,dy\\
&=-2\int_0^{\pi/4} \frac{y^2\ln\left(1-\cos2y\right)}{1+\cos2y}\,dy\\
&=-\frac{1}{4}\int_0^{\pi/2} \frac{t^2\ln\left(1-\cos t\right)}{1+\cos t}\,dt\qquad\Rightarrow\qquad t=2y\\
\end{align}
Use integration by parts by taking $u=t^2$ and $dv=\dfrac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt$, then
\begin{align}
v&=\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
\end{align}
Use integration by parts by taking $u=\ln\left(1-\cos t\right)$ and $dv=\dfrac{dt}{1+\cos t}$, by Weierstrass substitution: $x=\tan\left(\dfrac{t}{2}\right)$ then
\begin{align}
v=\int\frac{dt}{1+\cos t}=\int \,dx=\tan\left(\frac{t}{2}\right)=\frac{\sin t}{1+\cos t}
\end{align}
Hence
\begin{align}
\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-\int\frac{\sin t}{1+\cos t}\cdot\frac{\sin t}{1-\cos t}\,dt\\
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t
\end{align}
and
\begin{align}
I&=-\frac{1}{4}\left[\frac{t^2\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^3\right]_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}\left[\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^2\right]\,dt\\
&=\frac{\pi^3}{32}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt-\frac{\pi^3}{48}\\
&=\frac{\pi^3}{96}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt
\end{align}
Consider
\begin{equation}
I(a)=\int_0^{\pi/2} \frac{t\sin t}{1+\cos t}\ln\left(1-a\cos t\right)\,dt
\end{equation}
so that $I(0)=0$ and
\begin{align}
I'(a)&=-\int_0^{\pi/2} \frac{t\sin t\cos t}{(1-a\cos t)(1+\cos t)}\,dt\\
&=\frac{1}{1+a}\int_0^{\pi/2} \left(\frac{t\sin t}{1+\cos t}-\frac{t\sin t}{1-a\cos t}\right)\,dt\\
\end{align}
Now consider
\begin{equation}
I(b)=\int_0^{\pi/2} \frac{t\sin t}{1+b\cos t}\,dt
\end{equation}
Use integration by parts by taking $u=t$ and $dv=\dfrac{\sin t}{1+b\cos t}\,dt$, then 
\begin{align}
I(b)&= \frac{t\ln(1+b\cos t)}{b}\bigg|_0^{\pi/2}-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt\\
&=-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{align}
Consider
\begin{equation}
J(b)=\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{equation}
so that $J(0)=0$ and
\begin{align}
J'(b)&=\int_0^{\pi/2} \frac{\cos t}{1+b\cos t}\,dt\\
&=\frac{1}{b}\int_0^{\pi/2} \left(1-\frac{1}{1+b\cos t}\right)\,dt\\
&=\frac{\pi}{2b}-\int_0^{\pi/2} \frac{dt}{1+b\cos t}\qquad\Rightarrow\qquad x=\tan\left(\frac{t}{2}\right)\\
&=\frac{\pi}{2b}-\int_0^{1} \frac{2}{1+b+(1-b)x^2}\,dx\qquad\Rightarrow\qquad x=\sqrt{\frac{1+b}{1-b}}\tan z\\
&=\frac{\pi}{2b}-\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\\
J(b)&=\frac{\pi}{2}\ln b-\int\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\,db\\
\end{align}
Again we use integration by parts by taking $u=\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)$ and $dv=\dfrac{2}{\sqrt{1-b^2}}$, we have
\begin{align}
J(b)&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{1-b}\sqrt{\frac{1-b}{1+b}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{\sqrt{1-b^2}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\frac{\arcsin^2 b}{2}
\end{align}
Therefore
\begin{align}
I'(a)&=\frac{\pi^2}{8(1+a)}-\frac{1}{a(1+a)}\left[\frac{\pi}{2}\ln (-a)-2\arctan\left(\sqrt{\frac{1+a}{1-a}}\right)\arcsin (-a)-\frac{\arcsin^2(-a)}{2}\right]\\
\end{align}
From this step, I give up. Perhaps someone else want to continue it. Be my guest...
</s>