<s>[INST] Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.
 [/INST] It is not necessary to use complex analysis or to use power series to compute. From @Random Variable, we have
$$ I'(a) =\int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}=\int_{-1}^1 \frac{1}{1+ax}d\arcsin x. $$
Let $x=\sin t$. Then
$$ I'(a) =\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin t}dt. $$
Let $u=\tan\frac{t}{2}$. Then
\begin{eqnarray}
I'(a)&=&2\int_{-1}^{1} \frac{1}{u^2+1+2au}du=2\int_{-1}^{1} \frac{1}{(u+a)^2+1-a^2}du\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}-\arctan\frac{-1+a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}+\arctan\frac{1-a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt{1-a^2}}.
\end{eqnarray}
But $I(0)=0$ and so $ I(a)=\pi\arcsin a. $
</s>