How to solve $\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}} dx$ Consider the integral
$$\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx$$
How to start integrating?
Any hint would be appreciated.

Using the substitution $ x = \sin(t) $
The integral simplifies to
$$\int \sin^5t \cdot \ln \left[\frac{1+ \sin(t)}{1-\sin(t)}\right] dt$$
Further simplifications yields:
$$ 2\int \sin^5t \cdot \ln[\sec(t)+\tan(t)] dt $$
we recognize $\ln[\sec(t)+\tan(t)]$ as the antiderivative of $\sec(t)$ which suggests integrating by parts with $u = \ln[\sec(t)+\tan(t)]$ and $dv = \sin^5t \cdot dt$.
