An arctan integral

Evaluate the integral:

$$\mathcal{J}=\int_0^1 \frac{\arctan x}{x \sqrt{1-x^2}}\, {\rm d}x$$

Solution

Let us consider the integral $\displaystyle \mathcal{J}(a)= \int_{0}^{1}\frac{\arctan ax}{x \sqrt{1-x^2}} \, {\rm d}x$. Differentiating with respect to $a$ once, we have that:

\begin{align*}
\frac{\mathrm{d} }{\mathrm{d} a} \mathcal{J}(a) &= \int_{0}^{1}\frac{\partial }{\partial a} \frac{\arctan ax}{x \sqrt{1-x^2}} \, {\rm d}x \\
 &= \int_{0}^{1}\frac{1}{1+a^2 x^2} \cdot \frac{1}{\sqrt{1-x^2}} \, {\rm d}x\\
 &\!\!\!\!\overset{x=\sin t}{=\! =\! =\! =\!} \int_{0}^{\pi/2} \frac{{\rm d}u}{1+a^2 \sin^2 u} \\
 &= \frac{\pi}{2\sqrt{1+a^2}}
\end{align*}

Thus:

$$\mathcal{J}(a)= \frac{\pi}{2}\ln \left ( a+\sqrt{1+a^2} \right )+c$$

For $a=0$ we have $\mathcal{J}(0)=0$, thus $c=0$. Now, the initial integral is obtained by setting $a=1$. Hence:

$$\int_0^1 \frac{\arctan x}{x \sqrt{1-x^2}}\, {\rm d}x=\frac{\pi \ln \left(1+\sqrt{2}\right)}{2}$$