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.

It's easier to take the derivative of both sides according to $a$, than to perform the integration:
$$\int_0^1 \frac{2}{1-a^2x^2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{\sqrt{1-a^2}}$$
The left hand side giving:
$$2 \text{arctanh}\left(\frac{\sqrt{a^2-1}}{\sqrt{1 - x^2}}x\right)\frac{1}{\sqrt{a^2-1}}+C$$
Which, when applying the limits, gives:
$$\frac{\pi}{\sqrt{1-a^2}}+C$$
as desired. Now all that needs to be done is compare at a single point to prove that the $C=0$, which you already have done.
