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.

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.
