Differentiation under the integral sign for $\int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx$ Hello I have a problem that is:
$$\int_0^1\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
I try use the following integral
$$ \int_0^1\frac{dy}{1+x^2y^2}= \frac{\arctan(x)}{x}$$ 
My question: if I can do $$\frac{\arctan(x)}{x\sqrt{1-x^2}}= \int_0^1\frac{dy}{(1+x^2y^2)(\sqrt{1-x^2})}$$ and solve but I note that the integral is more difficult.
Any comment or any help will be well received.
Thanks.

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By following the user hint, we'll have:
\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 y \over 1 + x^{2}y^{2}}}^{\ds{\arctan\pars{x}/x}}\
\,\dd x =
\int_{0}^{1}\int_{0}^{1}{\dd x \over \root{1 - x^{2}}\pars{y^{2}x^{2} + 1}}
\,\dd y
\end{align}
With $x \to 1/x$, the above expression becomes
\begin{align}
\color{#f00}{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x} & =
\int_{0}^{1}
\int_{1}^{\infty}{x\,\dd x \over \root{x^{2} - 1}\pars{x^{2} + y^{2}}}\,\dd y\
\\[3mm] & \stackrel{x^{2}\ \to\ x}{=}\
\half\int_{0}^{1}
\int_{1}^{\infty}{\dd x \over \root{x - 1}\pars{x + y^{2}}}\,\dd y\
\\[3mm] & \stackrel{x - 1\ \to\ x}{=}\
\half\int_{0}^{1}
\int_{0}^{\infty}{\dd x \over \root{x}\pars{x + 1 + y^{2}}}\,\dd y
\\[3mm] & \stackrel{x\ \to\ x^{2}}{=}\
\int_{0}^{1}\int_{0}^{\infty}{\dd x \over x^{2} + y^{2} + 1}\,\dd y =
\int_{0}^{1}{1 \over \root{y^{2} + 1}}\ \overbrace{%
\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{\pi/2}}\ \,\dd y
\end{align}
With $y \equiv \sinh\pars{\theta}$, the last expression is reduced to
\begin{align}
\color{#f00}{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x} & =
{\pi \over 2}\int_{0}^{\textrm{arcsinh}\pars{1}}\,\dd\theta =
\color{#f00}{{\pi \over 2}\,\textrm{arcsinh}\pars{1} =
             {\pi \over 2}\,\ln\pars{\!\root{2} + 1}} \approx 1.3845
\end{align}
