How to evaluate the following definite integral $\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx$? 
$$\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx$$

My working is as follows;
Let 
$$I(a)=\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx$$
and thus 
$$I'(a)=0-0+\int_0^1\frac{x}{(1+a^2x^2)(x\sqrt{1-x^2})}dx$$
(I did this using Leibniz-Newton's formula. I took the partial derivative of the integrand with respect to $a$, and since the two limits of the integral are constants, the first two terms are zero.)
Now, I'm stuck... I've been trying this one for the last hour. Can someone hint me towards the right answer? (which I looked up to be $\frac{\pi}{2}\sinh^{-1}a$)
P.S: I am aware of this site's norm of keeping MathJax out of the titles, but I'm getting a message that a question with this title already exists... Can someone please help?
 A: By Feynman's trick
$$ \int_{0}^{1}\frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx = \int_{0}^{a}\int_{0}^{1}\frac{dx}{(1+b^2 x^2)\sqrt{1-x^2}}\,db = \int_{0}^{a}\frac{\pi}{2\sqrt{b^2+1}}\,db = \color{red}{\frac{\pi}{2}\text{arcsinh}(a).}$$
A: Now you need to solve 
$$I=I'(a)=0-0+\int_0^1\frac{x}{(1+a^2x^2)(x\sqrt{1-x^2})}dx$$
$$I=I'(a)=\int_0^1\frac{1}{(1+a^2x^2)(\sqrt{1-x^2})}dx$$
Substitue $x=\sin\left(u\right)$ thus $\mathrm{d}x=\cos\left(u\right)\,\mathrm{d}u$
$$I={\displaystyle\int}\dfrac{\cos\left(u\right)}{\sqrt{1-\sin^2\left(u\right)}\left(a^2\sin^2\left(u\right)+1\right)}\,\mathrm{d}u$$
$$I={\displaystyle\int}\dfrac{1}{a^2\sin^2\left(u\right)+1}\,\mathrm{d}u$$
Use $$\sin\left(u\right)=\dfrac{\tan\left(u\right)}{\sec\left(u\right)}$$
$$\sec^2\left(u\right)=\tan^2\left(u\right)+1$$
$$I={\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\sec^2\left(u\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\left(a^2+1\right)\tan^2\left(u\right)+1}}}\,\mathrm{d}u$$
Substitute $v=\tan\left(u\right)$ and $\mathrm{d}v=\dfrac{1}{\sqrt{a^2+1}}\,\mathrm{d}w$
$$I={\displaystyle\int}\dfrac{1}{\left(a^2+1\right)v^2+1}\,\mathrm{d}v$$
Substitue $w=\sqrt{a^2+1}v$
$$I=\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{\sqrt{a^2+1}}}}{\displaystyle\int}\dfrac{1}{w^2+1}\,\mathrm{d}w$$
$$I=\dfrac{\arctan\left(w\right)}{\sqrt{a^2+1}}$$
Putting back all the values you get
$$I=\dfrac{\arctan\left(\frac{\sqrt{a^2+1}x}{\sqrt{1-x^2}}\right)}{\sqrt{a^2+1}}+C$$
