Inverse Trigonometric Integrals How to calculate the value of the integrals
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx
$$
and
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$
 A: Here is a simple and a nice way to evaluate the first and the second integral.
Evaluation of $1^{\mbox{st}}$ Integral :
Making substitution $x=\tan\theta\,$ followed by integration by parts, we get
\begin{align}
\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx&=\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\\
&=-\theta^2\cot\theta\bigg|_0^{\Large\frac{\pi}{4}}+2\int_0^{\Large\frac{\pi}{4}}\theta\cot\theta\,d\theta\tag{1} \\
&=-\frac{\pi^2}{16}+2\theta\ln(\sin\theta)\bigg|_0^{\Large\frac{\pi}{4}}-2\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta\tag{2}\\
&=-\frac{\pi^2}{16}-\frac{\pi}{4}\ln2+G+\frac{\pi}{2}\ln2\tag{3}\\
&=\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2}
\end{align}

Evaluation of $2^{\mbox{nd}}$ Integral :
Again making substitution $x=\tan\theta\,$ followed by integration by parts, we get
\begin{align}
\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx&=\int_0^{\Large\frac{\pi}{4}}\frac{\theta^3\cos\theta}{\sin^3\theta}\,d\theta\\
&=-\left.\frac{\theta^3}{2\sin^2\theta}\right|_0^{\Large\frac{\pi}{4}}+\frac{3}{2}\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\tag{4}\\
&=-\frac{\pi^3}{64}++\frac{3}{2}\left[\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2}\right]\\
&=\color{blue}{\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2}
\end{align}
Explanation :
$(1)$ Integration by parts, $u=\theta^2\,\mbox{ and }\,dv=\dfrac{d\theta}{\sin^2\theta}$
$(2)$ Integration by parts, $u=\theta\,\mbox{ and }\,dv=\cot\theta\,d\theta$
$(3)$ The evaluation of $\displaystyle\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta$. See Mr. Tunk-Fey's answer, his answer is the best!
$(4)$ Integration by parts, $u=\theta^3\,\mbox{ and }\,dv=\dfrac{\cos\theta}{\sin^3\theta}\,d\theta$
Done! $\,$ (>‿◠)✌
A: The first integral is not that difficult to evaluate.  Note that
$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_1^{\infty} dx \frac{\arctan^2{x}}{x^2}\\ &=  \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2\end{align}$$
The first integral may be evaluated by a simple substitution $x=\tan{u}$ to get
$$\int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} = \int_0^{\pi/2} du \frac{u^2}{\sin^2{u}} $$
The latter integral is equal to $\pi \log{2}$; the derivation of this result may be found here.
The second integral is evaluated by expansion and repeated integration by parts, as follows:
$$\begin{align} \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2 &= \frac{\pi^2}{4} - \pi \int_0^1 dx \, \arctan{x} + \int_0^1 dx \, \arctan^2{x}\end{align} $$
The first integral on the RHS is 
$$ \begin{align}\int_0^1 dx \, \arctan{x} &= \left [ x \arctan{x} \right ]_0^1 - \int_0^1 dx \frac{x}{1+x^2}\\ &= \frac{\pi}{4} - \frac12 \log{2} \end{align} $$
The second integral is a little more involved, but it is along similar lines:
$$\begin{align} \int_0^1 dx \, \arctan^2{x} &=  \left [ x \arctan^2{x} \right ]_0^1 - 2 \int_0^1 dx \frac{x}{1+x^2} \arctan{x} \\ &= \frac{\pi^2}{16} - \left [\log{(1+x^2)} \arctan{x} \right ]_0^1 + \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} \end{align} $$
The latter integral on the RHS may be evaluated by recognizing that
$$\begin{align} \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} &= \int_0^1  dx \frac{\log{(1+x^2)}}{1+x^2} + \int_1^{\infty}  dx \frac{\log{(1+x^2)}}{1+x^2} \\ &= \int_0^1  dx \frac{\log{(1+x^2)}}{1+x^2} + \int_0^1  dx \frac{\log{(1+x^2)}- 2 \log{x}}{1+x^2}\end{align} $$
The latter integrals are obtained through a mapping $x \mapsto 1/x$ in the previous integral.  Thus,
$$\begin{align}\int_0^1  dx \frac{\log{(1+x^2)}}{1+x^2} &= \frac12 \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} +  \int_0^1  dx \frac{\log{x}}{1+x^2} \\ &= - \int_0^{\pi/2} du \, \log{\cos{u}} - G \\ &= \frac{\pi}{2} \log{2} - G  \end{align} $$
where $G$ is Catalan's constant.  The source of the first integral on the RHS is the same as that found in the above link (here).
Putting this all together, we get that
$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \pi \log{2} - \frac{\pi^2}{4} + \frac{\pi^2}{4} - \frac{\pi}{2} \log{2} - \frac{\pi^2}{16} + \frac{\pi}{4} \log{2} - \frac{\pi}{2} \log{2} + G \\ &= G + \frac{\pi}{4} \log{2} - \frac{\pi^2}{16} \end{align}$$
which matches up with other people's assertions.
A: 
How to calculate the value of the integrals?

Using Wolfram|Alpha Pro, one may obtain
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx=G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2$$
and
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx=\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2$$
Sorry for the Cleo's style answers, but the answer style is just like the OP style.
