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!

I thought it might be instructive to present two approaches that begin with the Feyman "trick" for differentiating under the integral.  We write
$$I(a)=\int_0^1\frac{\arctan (ax)}{x\sqrt{1-x^2}}\,dx$$ 
Then, we differentiate with $I(a)$ to find
$$I'(a)=\int_0^1\frac{1}{(1+x^2a^2)\sqrt{1-x^2}}\,dx$$
can be evaluated by first substituting $x=\sin u$ so that 
$$\begin{align}
I'(a)&=\int_0^{\pi/2}\frac{1}{1+a^2\sin^2 u}\,du\\\\
&=\frac12 \int_{-\pi/2}^{\pi/2}\frac{1}{1+a^2\sin^2 u}\,du
\end{align}$$
Next, we use the trigonometric identity $\sin^2u=\frac{1-\cos(2u)}{2}$ so that
$$\begin{align}
I'(a)&=\int_0^{\pi/2}\frac{1}{1+\frac12a^2-\frac12a^2\cos (2u)}\,du \tag 1\\\\
&=\frac14 \int_{-\pi}^{\pi}\frac{1}{1+\frac12a^2-\frac12a^2\cos (u)}\,du \tag 2
\end{align}$$
We pursue evaluation of $(1)$ using the Weirestrass Substitution and evaluation of $(2)$ using contour integration.

First, we enforce the substitution $2u=\tan(x/2)$ in $(1)$.  Then, we obtain
$$\begin{align}
I'(a)&=2\int_{0}^{\infty}\frac{1}{1+4(1+a^2)x^2}\,dx\\\\
&=\frac{2}{2\sqrt{1+a^2}}\left.\arctan\left(2\sqrt{1+a^2}\,x\right)\right|_{0}^{\infty}\\\\
&=\frac{\pi}{2\sqrt{1+a^2}}
\end{align}$$

Alternatively, we let $u=e^{iz}$ in $(2)$ and write
$$\begin{align}
I'(a)&=\frac{i}{2a^2}\oint_{|z|=1}\frac{1}{z^2-\left(1+\frac{2}{a^2}\right)z+1}\,dz\\\\
&=\frac{i}{a^2}(2\pi i) \,\,\text{Res}\left(\frac{1}{z^2-2\left(1+\frac{2}{a^2}\right)z+1},z=\left(1+\frac{2}{a^2}\right)-\frac{2\sqrt{1+a^2}}{a^2}\right)\\\\
&=\pi/2\sqrt{1+a^2}
\end{align}$$

Finally, following the approach used by @Manu we find from $I'(a)$, $I(1)$ is 
$$I(1)=\pi \log(1+\sqrt{2})$$
And we are done!
