Using differentiation for computing $\int_{0}^{1}\frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx$ How can I prove that I can differentiate the integral:
$$ \int_{0}^{1} \frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx $$
First I have to prove this integral converges.
Next I have to prove the integral $$ \int_{0}^{1} f_{a}'(x, a)\,dx $$ converges uniformly.
Then I can differentiate integral with Leibniz's rule.
I tried Weierstrass and Dirichlet's tests. Nothing succeeded. 
Thank you for help in advance.

The given hints contain pretty much everything. $\frac{d}{da}\arctan(ax)=\frac{x}{1+a^2 x^2}$ and 
$$ \int_{0}^{1}\frac{dx}{(1+a^2 x^2)\sqrt{1-x^2}} \stackrel{x\mapsto\sin\theta}{=} \int_{0}^{\pi/2}\frac{d\theta}{1+a^2\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+a^2\cos^2\theta}\\\stackrel{\theta\mapsto\arctan u}{=}\int_{0}^{+\infty}\frac{du}{(1+a^2)+u^2}=\frac{\pi}{2\sqrt{1+a^2}} $$
so by the dominated/monotone convergence theorem
$$ \int_{0}^{1}\frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx = \frac{\pi}{2}\int_{0}^{a}\frac{dv}{\sqrt{1+v^2}}=\color{blue}{\frac{\pi}{2}\text{arcsinh}(a)}.$$
Differentiation under the integral sign is also known as Feynman's trick: it is pretty efficient in dealing with integrals involving $\log$ or $\arctan$ or $\text{arctanh}$.
