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 $a=\tan s$. Under $x\to\sin x$, one has
\begin{eqnarray}
I(s)&:=&\int_0^1\frac{\tan^{-1}(ax)}{x\sqrt{1-x^2}}\,dx\\
&=&\int_0^{\pi/2}\frac{\tan^{-1}(\tan s\sin x)}{\sin x}\,dx.
\end{eqnarray}
Then
\begin{eqnarray}
I'(s)&=&\int_0^{\pi/2}\frac{\sec^2s}{1+\tan^2s\sin^2 x}\,dx\\
&=&\sec^2s\int_0^{\pi/2}\frac{1}{1+\tan^2s\sin^2 x}\,dx\\
&=&\sec^2s\cdot\frac{\pi}{2}\cos s\\
&=&\frac{\pi}{2}\sec s.
\end{eqnarray}
So
$$ I(s)-I(0)=\frac{\pi}{2}\int_0^s\frac{1}{\cos t}dt=\frac{\pi}2\ln(\sec s+\tan s) $$
or
$$ I=\frac{\pi}{2}\ln(a+\sqrt{1+a^2}).$$
