Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ Wanting to calculate the integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ it will certainly already known to many of you that an interesting way to attack it is to refer to the method of integration and differentiation with respect to a parameter, getting $\frac{\pi}{2}\,\log\left(1+\sqrt{2}\right)$. 
Instead, what it does not at all clear is how software such as Wolfram Mathematica can calculate that result in an exact manner and not only approximate. Can someone enlighten me? Thanks!

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\begin{align}
&\color{#f00}{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x} =
\int_{0}^{1}{1 \over \root{1 - x^{2}}}\
\overbrace{\int_{0}^{1}{\dd t \over 1 + x^{2}t^{2}}}
^{\ds{\arctan\pars{x} \over x}}\ \,\dd x
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}{\dd x \over \root{1 - x^{2}}\pars{1 + t^{2}x^{2}}}
\,\dd t
\\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,&
\int_{0}^{1}
\int_{1}^{\infty}{x\,\dd x \over \root{x^{2} - 1}\pars{x^{2} + t^{2}}}\,\dd t
\label{1}\tag{1}
\end{align}

Note that the last substitution $\ds{\pars{~x \mapsto 1/x~}}$ leads to a $\ds{\ul{trivial\ integration}}$.


With the substitution $\ds{x^{2} \mapsto x}$ in the last integration $\ds{~\pars{\mbox{see}\ \eqref{1}}~}$,
\begin{align}
&\color{#f00}{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x} =
\half\int_{0}^{1}
\int_{1}^{\infty}{\dd x \over \root{x - 1}\pars{x + t^{2}}}\,\dd t
\\[5mm] &
\stackrel{x\ \equiv\ 1 + y^{2}}{=}\,\,\,\
\int_{0}^{1}\int_{0}^{\infty}{\dd y \over y^{2} + 1 + t^{2}}\,\dd t =
\int_{0}^{1}{1 \over \root{1 + t^{2}}}
\int_{0}^{\infty}{\dd y \over y^{2} + 1}\,\dd t
\\[5mm] = &\ 
{\pi \over 2}\int_{0}^{1}{\dd t \over \root{1 + t^{2}}}
\stackrel{t\ =\ \sinh\pars{\theta}}{=}\,\,\,
{\pi \over 2}\int_{0}^{\mrm{arcsinh}\pars{1}}\,\dd\theta =
\color{#f00}{{\pi \over 2}\,\mrm{arcsinh}\pars{1}}
\\[5mm] = &\
\color{#f00}{{\pi \over 2}\,\ln\pars{1 + \root{2}}}
\end{align}
