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

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.
