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.

We will use Taylor series of arctan, so we have :
$$f(a)=\int_0^1\frac{\arctan ax}{x\sqrt{1-x^2}}=\sum_{n=0}^\infty{\frac{(-1)^na^{2n+1}}{2n+1}}\biggl(\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\biggr)$$
Then we use integration by substitution for calculate the given integral (by choosing $x=\sin t$) :
 $$\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}=\int_0^{\pi/2}\sin^{2n}t=W_{2n}$$
And we know calculate wallis'integral, 
$$W_{2n}=\frac{\pi}{2}\frac{(2n)!}{2^{2n}(n!)^2}$$
So we have :
$$\int_0^1\frac{\arctan ax}{x\sqrt{1-x^2}}=\frac{\pi}{2}\sum_{n=0}^\infty{\frac{(-1)^n(2n)!}{4^n(2n+1)(n!)^2}a^{2n+1}}$$
Next step we will use taylor series of arcsin 
(see this link https://fr.wikipedia.org/wiki/S%C3%A9rie_de_Taylor), so we have :
 $$f(a)=\frac{\pi}{2}\frac{\arcsin( {ia})}{{i}}=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|$$
