Evaluate $\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx$ Evaluate $$\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx\,,$$  where $a$ being parameter. I am not able to solve this.

Let $$I(a) = \int_0^{1}\frac{\arctan {ax}}{x\sqrt{1-x^2}}dx$$
Differentiating the integral with respect to $a$,
$$I'(a) = \int_0^{1}\frac{dx}{\sqrt{1-x^2}(1+a^2x^2)}$$
Let, $$u^2 = \frac{1-x^2}{1+a^2x^2} \implies x^2=\frac{1-u^2}{1+a^2u^2} \therefore xdx=-\frac{u(1+a^2)}{(1+a^2u^2)^2}du$$
Therefore, the integral reduces to,
$$I'(a)=\frac{1}{\sqrt{1+a^2}}\int_0^{1}\frac{du}{\sqrt{1-u^2}}$$
$$\therefore I'(a)=\frac{\pi}{2\sqrt{1+a^2}} \implies I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|+C$$ where C is an undetermined constant that can be found by putting $a=0$ in the original integral which evaluates to $0$. Therefore, $C=0$ and the integral is 
$$I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|$$
