Finding $ \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$ 
Finding $\displaystyle \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$

Try: Let $$\displaystyle I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$ (Function is even )
$$I = 2\int^{1}_{0}\bigg[\ln(1+x)-\ln(1-x)\bigg]\frac{x^3}{\sqrt{1-x^2}}dx$$ 
Using By parts
$$I = 2\bigg[\ln\bigg(\frac{1+x}{1-x}\bigg)\bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)\bigg]\bigg|^{1}_{0}+4\int^{1}_{0}\frac{1}{1-x^2}\cdot \bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)dx$$
Could some help me to solve second Integration, Thanks
 A: Since $\int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac{1}{3}(2+x^2)\sqrt{1-x^2}$, integration by parts yields
\begin{align*}
\int_{-1}^{1} \frac{x^3}{\sqrt{1-x^2}} \log\bigg(\frac{1+x}{1-x}\bigg) \, dx
&= \int_{-1}^{1} \frac{2(2+x^2)}{3\sqrt{1-x^2}} \, dx.
\end{align*}
Now the substitution $x = \sin\theta$ quickly yields the answer $\frac{5}{3}\pi$.
A: Let us consider your integral:
$$I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$
Separate the integral: 
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx-\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1-x)\ dx$$
Consider the right-hand integral. Let $x=-t$ such that $dx=-dt$, where $t\in (1,-1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx-\int^{-1}_{1}\frac{(-t)^3}{\sqrt{1-(-t)^2}}\ln (1-(-t))\ (-dt)$$
Incorporate a minus sign into the integral in order to switch to interval to $t\in (-1,1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx+\int^{1}_{-1}\frac{t^3}{\sqrt{1-t^2}}\ln (1+t)\ dt$$
Recognize that $t$ is a dummy variable. Let $t=x$ such that $dt=dx$, where $x\in (-1,1)$:
$$I= \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx+\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx=2\int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln (1+x)\ dx$$
Apply integration by parts:
$$I=-\frac{2}{3}\sqrt{1-x^2}(x^2+2)\ln(1+x)\bigg|^{1}_{-1}+\frac{2}{3}\int^{1}_{-1}\sqrt{1-x^2}\frac{x^2+2}{1+x}\ dx=\frac{2}{3}\int^{1}_{-1}\sqrt{1-x^2}\frac{x^2+2}{1+x}\ dx$$
Let $x=\sin(u)$ such that $dx=\cos(u)\ du$, where $u\in (-\pi/2,\pi/2]$:
$$I=\frac{2}{3}\int^{\pi/2}_{-\pi/2}\cos^2(u)\frac{\sin^2(u)+2}{1+\sin(u)}\ du=\frac{2}{3}\int^{\pi/2}_{-\pi/2}(\sin^2(u)+2)\frac{1-\sin^2(u)}{1+\sin(u)}\ du$$
Perform long-division on the integrand:
$$I=\frac{2}{3}\int^{\pi/2}_{-\pi/2}(\sin^2(u)+2)(1-\sin(u))\ du=\frac{2}{3}\int^{\pi/2}_{-\pi/2}\sin^2(u)\ du-\frac{2}{3}\int^{\pi/2}_{-\pi/2}\sin^3(u)\ du+\frac{4}{3}\int^{\pi/2}_{-\pi/2}du-\frac{4}{3}\int^{\pi/2}_{-\pi/2}\sin(u)\ du$$
Since $\sin^3(u)$ and $\sin(u)$ are odd, their integrals evaluate to zero. Rewrite $\sin^2(u)$ by making use of a double-angle identity:
$$I=\frac{1}{3}\int^{\pi/2}_{-\pi/2}(1-\cos(2u))\ du+\frac{4}{3}\int^{\pi/2}_{-\pi/2}du=\frac{5}{3}u-\frac{1}{6}\sin(2u)\bigg|^{1}_{-1}$$
Thus,
$$I=\frac{5}{3}\pi$$
