<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] The integral in question,
\begin{align}
I = \int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}}
\end{align}
can be separated into the two integrals
\begin{align}
I = \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx - \int_{0}^{1} \frac{ \ln(1-ax)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
which will be labled $I_{1}$ and $I_{2}$. Now
\begin{align}
I_{1} &= \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} a^{n}}{n} \ \int_{0}^{1} \frac{x^{n-1} \ dx}{\sqrt{1-x^{2}}} \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-a)^{n}}{n} \ B(n/2, 1/2) \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-2a)^{n} \ \Gamma^{2}(n/2)}{n!} \\
&= - \frac{1}{4} \left[ \sum_{k=0}^{\infty} \Gamma^{2}(k+1/2) \frac{(-2a)^{2k+1}}{(2k+1)!} + \sum_{k=0}^{\infty} \frac{(k!)^{2} (-2a)^{2k+2}}{(2k+2)!} \right] \\
&= - \frac{1}{4} \left[ -2\pi a \sum_{k=0}^{\infty} \frac{(1/2)_{k} (1/2)_{k} a^{2k}}{ k! (3/2)_{k}} + 4 a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}} \right] \\
&= \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
In a similar manor, 
\begin{align}
I_{2} &= - \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
Since $I = I_{1} - I_{2}$ then 
\begin{align}
\int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}} = \pi \ \sin^{-1}(a)
\end{align}
which is the desired value. 
</s>