How to solve $\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}} dx$

Question

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.

Answer 1

$$ \int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx $$ $$ u = \ln\frac{x+1}{1-x} \quad dv = \frac{x^5 dx}{\sqrt{1-x^2}} $$ $$ v = \int \frac{x^5 dx}{\sqrt{1-x^2}} \quad x = \sin\theta \quad v = \int \sin^5\theta d\theta = -\int (1 - 2\cos^2\theta + \cos^4\theta ) d\cos\theta = \\ - \cos\theta + 2/3\cos^3\theta - 1/5\cos^5\theta = -\sqrt{1-x^2}+2/3(1-x^2)^{3/2} - 1/5 (1-x^2)^{5/2} $$ $$ du = \frac{1-x}{1+x} \frac{2dx}{(1-x)^2} = du = \frac{2dx}{(1+x)(1-x)} $$ $$ \int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx = (\ln\frac{x+1}{1-x})(-\sqrt{1-x^2}+2/3(1-x^2)^{3/2} - 1/5 (1-x^2)^{5/2}) - \int [(-\sqrt{1-x^2}+2/3(1-x^2)^{3/2} - 1/5 (1-x^2)^{5/2})\frac{2dx}{(1+x)(1-x)}] $$ $$ \int [(-\sqrt{1-x^2}+2/3(1-x^2)^{3/2} - 1/5 (1-x^2)^{5/2})\frac{2dx}{(1+x)(1-x)}] = \int [(-(1-x^2)^{-1/2}+2/3(1-x^2)^{1/2} - 1/5 (1-x^2)^{3/2})2dx] $$ $$ \int [(-(1-x^2)^{-1/2}dx = -\arcsin(x) $$ $$ \int 2/3(1-x^2)^{1/2} dx = 2/3 [\frac{1}{2}\left(\arcsin \left(x\right)+\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right)\right)] $$ $$ \int 1/5 (1-x^2)^{3/2} dx = 1/5[x\left(1-x^2\right)^{\frac{3}{2}}+\frac{3}{8}\left(\arcsin \left(x\right)-\frac{1}{4}\sin \left(4\arcsin \left(x\right)\right)\right)] $$

All the integrals involved have been evaluated. Now you need to sum :)

Answer 2

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 $dv = \sin^5t \cdot dt$.