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

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 $\frac{5}{3}\pi$.
