Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate 
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$  
where $C$ is Catalan's constant. 
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}\,dx$$
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$\ds{\int_{0}^{1}\arctan^{2}\pars{x}\,\dd x
     ={\pi^{2} \over 16} + {\pi\ln\pars{2} \over 4} - C:\ {\large ?}.\quad}$
$\ds{C}$ is the Catalan Constant.

Set $\ds{\quad x \equiv \tan\pars{\theta}\quad\imp\quad\theta = \arctan\pars{x}}$:
  \begin{align}&\color{#c00000}{\int_{0}^{1}\arctan^{2}\pars{x}\,\dd x}
=\int_{0}^{\pi/4}\theta^{2}\sec^{2}\pars{\theta}\,\dd\theta
={\pi^{2} \over 16} - \int_{0}^{\pi/4}\tan\pars{\theta}\pars{2\theta}\,\dd\theta
\\[3mm]&={\pi^{2} \over 16} + \ln\pars{1 \over \root{2}}2\,{\pi \over 4} -2\int_{0}^{\pi/4}\ln\pars{\cos\pars{\theta}}\,\dd\theta
\\[3mm]&={\pi^{2} \over 16} - {1 \over 4}\,\pi\ln\pars{2} - \int_{0}^{\pi/4}
\ln\pars{\half\bracks{2\sin\pars{\theta}\cos\pars{\theta}}\cot\pars{\theta}}
\,\dd\theta
\\[3mm]&={\pi^{2} \over 16} - {1 \over 4}\,\pi\ln\pars{2}
+ {1 \over 4}\,\pi\ln\pars{2} - \int_{0}^{\pi/4}\ln\pars{\cot\pars{\theta}}\,\dd\theta
-\int_{0}^{\pi/4}\ln\pars{\sin\pars{2\theta}}\,\dd\theta
\end{align}

However, a integral representation of $\ds{C}$ is given by:
$$
C = \int_{0}^{\pi/4}\ln\pars{\cot\pars{\theta}}\,\dd\theta
$$
such that
$$
\color{#c00000}{\int_{0}^{1}\arctan^{2}\pars{x}\,\dd x}
={\pi^{2} \over 16} - C
-\half\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}\tag{1}
$$

Also,
  \begin{align}&\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
=\half\bracks{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta
+\int_{0}^{\pi/2}\ln\pars{\cos\pars{\theta}}\,\dd\theta}
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{\sin\pars{2\theta} \over 2}\,\dd\theta
=-\,{1 \over 4}\,\pi\ln\pars{2}
+{1 \over 4}\int_{0}^{\pi}\ln\pars{\sin\pars{\theta}}\,\dd\theta
\\[3mm]&=-\,{1 \over 4}\,\pi\ln\pars{2}
+{1 \over 4}\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta
+{1 \over 4}\int_{\pi/2}^{\pi}\ln\pars{\sin\pars{\theta}}\,\dd\theta
\\[3mm]&=-\,{1 \over 4}\,\pi\ln\pars{2}
+\half\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
\end{align}

$$
\imp\quad\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta} = -\,\half\,\pi\ln\pars{2}
$$

Replace this result in $\pars{1}$:
  $$\color{#66f}{\large%
\int_{0}^{1}\arctan^{2}\pars{x}\,\dd x
={\pi^{2} \over 16} + {\pi\ln\pars{2} \over 4} - C} \approx 0.2453
$$

A: If one know that  $\displaystyle \int_0^{\dfrac{\pi}{2}}\log (\sin (x))dx=-\dfrac{\pi}{2}\log(2)$
so then :
$\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx=\dfrac{\pi}{2}\log(2)$   (by integration by parts)
Let $I=\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx$
Change of variable $u=\tan x$ :
$I=\displaystyle \int_0^{+\infty} \dfrac{\arctan x}{x(1+x^2)}dx$
$I=\displaystyle \int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty} \dfrac{\arctan x}{x(1+x^2)}dx$
in the second integral, in the right member make the change of variable $u=\dfrac{1}{x }$
$I= \displaystyle\int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan
\Big(\dfrac{1}{x}\Big) }{1+x^2}dx$
For $x>0$ one have $\arctan\Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$
$I= \displaystyle\int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
But:
$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$
So then:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-\displaystyle\int_0^1 \dfrac{x\arctan x}{1+x^2}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
Hence:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-2\displaystyle\int_0^1 \dfrac{x\arctan x}{1+x^2}dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$
and so:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-2\displaystyle\int_0^1 \dfrac{x\arctan x
}{1+x^2}dx+\dfrac{\pi}{4}\log(2)$
But:
the derivative of $x\rightarrow(\arctan x)^2$ is $x\rightarrow\dfrac{2\arctan x }{1+x^2}$
So then:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-\Big(\big[ x(\arctan x)^2\big]_0^1-\int_0^1(\arctan x)^2 dx\Big)+\dfrac{\pi}{4}\log(2)$
$I=\displaystyle\int_0^1\dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1(\arctan x)^2 dx$
But $I=\dfrac{\pi}{2}\log(2)$ so then:
$$\displaystyle\int_0^1 (\arctan x)^2
dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\log(2)$$
where $\displaystyle G=\int_0^1 \dfrac{\arctan x}{x}dx$ is the Catalan's constant.
A: I believe it is interesting to look at your definite integral in terms of the more general result
$$\begin{align} \int (\arctan x)^2dx=&x(\arctan x)^2 -\frac{1}{2}\log(1+x^2)\arctan x+ \log2\arctan x+\tag{1}\label{eq:main}\\
&-\Im\left\{\mbox{Li}_2\left(\frac{1+ix}{2}\right)\right\}+K.\end{align}$$
where $\mbox{Li}_2(z)$ is dilogarithm function. The above expression can be calculated by letting
$$\omega = \frac{1+ix}{2}=|\omega|\mbox{e}^{i\arctan x}\tag{2}\label{eq:omega}$$
and then noting that
$$|\omega|^2 = \omega \cdot (1-\omega).\tag{3}\label{eq:mod}$$
Thus, from \eqref{eq:omega}, 
$$\begin{align} \log \omega &= \log|\omega|+i\arctan x=\\
&=\frac{1}{2}\log|\omega|^2+i\arctan x=\\
&\stackrel{\eqref{eq:mod}}=\frac{1}{2}\log\omega + \frac{1}{2}\log(1-\omega)+ i\arctan x.\end{align}$$
This finally gives the following expression for the inverse tangent function:
$$\arctan x = -\frac{i}{2}[\log \omega + \mbox{Li}_1(\omega)],\tag{4}\label{eq:arctan}$$
where 
$$ \mbox{Li}_1(z) = -\log(1-z).$$
Using \eqref{eq:arctan} to calculate the integral,
$$\begin{align}\int(\arctan x)^2dx &= -\frac{1}{4}\int[\log\omega + \mbox{Li}_1(\omega)]^2 dx=\\
&=\frac{i}{2}\int[\log\omega +\mbox{Li}_1(\omega)]^2d\omega,\end{align}$$
and recalling that the integral definition of the dilogarithm is
$$ \mbox{Li}_2(z) = -\int_0^z\frac{\log(1-u)}{u}du = \int_0^z\frac{\mbox{Li}_1(u)}{u}du$$
easily leads to \eqref{eq:main}.
So the definite integral is
$$\int_0^1 (\arctan x)^2 dx = \frac{\pi^2}{16} + \frac{(\log 2) \pi}{8}- \Im\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\}.$$
The value of the dilogarithm in $\frac{1+i}{2}$ can be determined using Landen's identity, as detailed here, which was pointed out to me by user153012:
$$
\mbox{Li}_2\left(\frac{1+i}{2}\right) = \frac{5\pi^2}{96} - \frac{(\log 2)^2}{8} + i\left[C - \frac{(\log 2)\pi}{8}\right],
$$
$C$ being Catalan's constant. This confirms the result obtained in the previous answers, i.e.
$$\int_0^1 (\arctan x)^2 dx = \frac{\pi^2}{16} + \frac{(\log 2) \pi}{4} - C $$
A: \begin{align}
\int_0^1 \arctan^2x dx\overset{x=\tan t}= &\int_0^{\frac\pi4}t^2\sec^2tdt \overset {IBP}=
{\pi^2\over16} - \int_0^{\frac\pi4}2 t\tan t dt\\
=& 
{\pi^2\over16} + \int_0^{\frac\pi4}t \>d[\ln(2\cos^2t)] dt 
 =
{\pi^2\over16} - \int_0^{\frac\pi4}\ln(2\cos^2t) dt\\
 =& 
{\pi^2\over16} - \int_0^{\frac\pi4}\ln\sin2t dt 
+ \int_0^{\frac\pi4}\ln\tan t dt \\
=&{\pi^2\over16} + \frac{\pi}{4} \ln2-C
\end{align}
A: Integrating by parts twice,
$$ \begin{align} \int_{0}^{1} (\arctan x)^{2} \ dx &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \ dx \end{align}$$
Let $x = \tan t $.
Then
$$\begin{align}\int_{0}^{1} (\arctan x)^{2} \ dx  &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2  - 2 \int_{0}^{\pi /4} \ln (\cos t) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\  &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - C \end{align}$$ 
Fourier series of Log sine and Log cos
A: First Approach - Using Euler sums
We begin by finding the Maclaurin series expansion for $\arctan^2 x$. 
Since
$$\arctan x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1}, \qquad |x| < 1,$$
applying the Cauchy product to the product between two inverse tangent functions leads to:
\begin{align}
\arctan^2 x &= \arctan x \cdot \arctan x\\
&= \left (\sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1} \right ) \cdot \left (\sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1} \right )\\
&= \sum_{n = 1} \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) x^{2n}, \qquad |x| \leqslant 1.
\end{align}
Here $H_n$ is the $n$th harmonic number.
Making use of this result, the integral becomes
\begin{align}
I &= \int_0^1 \arctan^2 x \, dx\\
&= \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n} \, dx\\
&= \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n (2n + 1)} \left (H_{2n} - \frac{1}{2} H_n \right )\\
&= \sum_{n = 1}^\infty (-1)^{n + 1} \left (\frac{1}{n} - \frac{2}{2n + 1} \right ) \left (H_{2n} - \frac{1}{2} H_n \right )\\
&= \frac{1}{2} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} - \sum_{n = 1}^\infty \frac{(-1)^n H_n}{2n + 1} - \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n} + 2 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{2n+1}.\tag1
\end{align}
Values for each of the above four alternating Euler sums can be found here. Theirs values are:

$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} = \frac{1}{2} \ln^2 2 - \frac{\pi^2}{12}, \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_n}{2n + 1} = \mathbf{G} - \frac{\pi}{2} \ln 2,$$
$$\sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n} = \frac{1}{4} \ln^2 2 - \frac{5\pi^2}{48}, \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{2n + 1} = -\frac{\pi}{8} \ln 2.$$

On substituting into (1) one obtains
$$\int_0^1 \arctan^2 x \, dx = \frac{\pi^2}{16} + \frac{\pi}{4} \ln 2 - \mathbf{G},$$
as desired.
Second Approach - Integral of a function and its inverse
Let $f(x) = \arctan^2 x$. As $f$ is a monotonically increasing function on the interval $[0,1]$ an inverse given by $f^{-1} (x) = \tan (\sqrt{x})$ exists on the interval $[0,\frac{\pi^2}{4})$. Making use of the result:
$$\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1} (x) \, dx = b \cdot f(b) - a \cdot f(a),$$
noting that for our $f$ we have $f(a) = f(0) = 0$ and $f(b) = f(1) = \frac{\pi^2}{16}$, one sees that
$$\int_0^1 \arctan^2 x \, dx + \int_0^{\frac{\pi^2}{16}} \tan (\sqrt{x}) \, dx = \frac{\pi^2}{16}.$$
Thus
\begin{align}
\int_0^1 \arctan^2 x \, dx &= \frac{\pi^2}{16} - \underbrace{\int_0^{\frac{\pi^2}{16}} \tan (\sqrt{x}) \, dx}_{x \, \mapsto \, x^2}\\
&= \frac{\pi^2}{16} - 2 \int_0^{\frac{\pi}{4}} x \tan x \, dx\\
&= \frac{\pi^2}{16} - 2 \left (-x \ln (\cos x) \Big{|}_0^{\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx \right ) \qquad (\text{by parts})\\
&= \frac{\pi^2}{16} - \frac{\pi}{4} \ln 2 - 2 \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx\\
&= \frac{\pi^2}{16} - \frac{\pi}{4} \ln 2 - 2 \left (\frac{\mathbf{G}}{2} - \frac{\pi}{4} \ln 2 \right )\tag1\\
&= \frac{\pi^2}{16} + \frac{\pi}{4} \ln 2 - \mathbf{G},
\end{align}
as expected. Note that in ($1$) we have made use of the result
$$\int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx = \frac{\mathbf{G}}{2} - \frac{\pi}{4} \ln 2,$$
several proofs for which can be found here.
A: To get rid of the inverse, we let $y=\arctan x$ and transform the integral into
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{4}} y^{2} \sec ^{2} y d y \\
&=\int_{0}^{\frac{\pi}{4}} y^{2} d(\tan y) \\
& \stackrel{IBP}{=} \left[y^{2} \tan y\right]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} y \tan y d y \\
&=\frac{\pi^{2}}{16}+2 \int_{0}^{\frac{\pi}{4}} y d(\ln (\cos y)) \\
&  \stackrel{IBP}{=}\frac{\pi^{2}}{16}+2[y \ln (\cos y)]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} \ln (\cos y) d y \\
&=\frac{\pi^{2}}{16}+2\left[\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)\right]-2\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right) \\
&=\frac{\pi^{2}}{16}+\frac{\pi}{4} \ln 2-G
\end{aligned}
$$
where the last integral comes from my post.
A: $$\begin{aligned}\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}dx & =\underbrace{ \int_{0}^{1} \frac{\ln \left(1+x^{2}\right)}{1+x^{2}} d x}_{J} - \underbrace{\int_{0}^{1} \frac{\ln x}{1+x^{2}} d x}_{K}\end{aligned}$$
For the integral $J$, letting $x=\tan x$ gives
$$
\begin{aligned}
J =\int_{0}^{\frac{\pi}{4}} \ln \left(\sec ^{2} \theta\right) d \theta &=-2 \int_{0}^{\frac{\pi}{4}} \ln (\cos \theta) d \theta =\frac{\pi}{2} \ln 2-G
\end{aligned}
$$
where the last integral comes from my post 1.
For the integral $K$, we consider the integral
$$
\begin{aligned}
K(a) &=\int_{0}^{1} \frac{x^{a}}{1+x^{2}} d x =\int_{0}^{1} \sum_{k=0}^{\infty}(-1)^{k} x^{2 k+a} d x =\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1+a}
\end{aligned}
$$
Then differentiating $K(a)$ w.r.t. $a$ at $a=0$ yields
$$
\begin{aligned}
K &=K^{\prime}(0) =\sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2 k+1)^{2}} =-G
\end{aligned}
$$
Therefore we can conclude that
$$ \boxed{I=\left(\frac{\pi}{2} \ln 2-G\right)+G=\frac{\pi}{2} \ln 2}$$
