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
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$

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! $\,$ (>‿◠)✌
