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.
 A: Why not to use the Taylor series of $\arctan(x)$?
$$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\sum_{n\geq 0}\frac{(-1)^n}{4n+2}\int_{0}^{1}z^{n-\frac{1}{2}}(1-z)^{-1/2}\,dx $$
by Euler's beta function the LHS is converted into:
$$ \sum_{n\geq 0}\frac{(-1)^n \Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{(4n+2)\Gamma(n+1)}=\frac{\pi}{2}\sum_{n\geq 0}\binom{-\frac{1}{2}}{n}\frac{(-1)^n}{2n+1}=\frac{\pi}{2}\int_{0}^{1}\frac{dx}{\sqrt{1+x^2}}$$
and we finally have:

$$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}\,dx = \frac{\pi}{2}\text{arcsinh}(1) = \color{red}{\frac{\pi}{2}\log(1+\sqrt{2})}.$$

An alternative approach is the following one. If we set
$$ I(k)=\int_{0}^{\pi/2}\frac{\arctan(k\sin\theta)}{\sin\theta}\,d\theta $$
our integral is $I(1)$ and $I(0)=0$, while:
$$ I'(k) = \int_{0}^{\pi/2}\frac{d\theta}{1+k^2\sin^2\theta} = \frac{\pi}{2\sqrt{1+k^2}}$$
through the substitution $\theta=\arctan t$. Now $I(1)=\int_{0}^{1}I'(k)\,dk$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\imp}{\Longrightarrow}
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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}
A: Ok let's star
$$I(y)=\int_{0}^{1}\frac{\arctan(yx)}{x\sqrt{1-x^2}}\,dx$$ for use the formula above then use differentiation under integral sign and integral methods.
