When I encountered this question, I just wondered whether there is a simple way to tackle the integral. After trying many substitutions, I still failed. Suddenly, integration by parts come to my mind and helped me solve the problem. I first integrate the part $$\dfrac{x^{3}}{\sqrt{1-x^{2}}}$$ and then evaluate it via integration by parts.
Are there any alternate methods other than integration by parts?
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Afterwards, I found that the substitution of trigonometric function works as well as integration by parts!
Letting $x=\cos 2 \theta$ yields$$ \begin{aligned} I &=\int_{\frac{\pi}{4}}^{0} \frac{\cos ^{3} 2 \theta \ln \left(\frac{2 \cos ^{3} \theta}{2 \sin ^{2} \theta}\right)}{\sqrt{1-\cos ^{3} 2 \theta}}(-2 \sin 2 \theta) d \theta \\ &=-4 \int_{0}^{\frac{\pi}{4}} \cos ^{3} 2 \theta \ln (\tan \theta) d \theta \end{aligned} $$
Noting that $$ \begin{aligned} \int \cos ^{3} 2 \theta d \theta &=\frac{1}{2} \int\left(1-\sin ^{2} 2 \theta\right) d(\sin 2 \theta) \\ &=\frac{1}{2}\left(\sin 2 \theta-\frac{\sin ^{3} 2 \theta}{3}\right)+C \end{aligned} $$
We can now use integration by parts again. $$ \begin{aligned} I=&-4\left(\left[\frac{1}{2}\left(\sin 2 \theta-\frac{\sin ^{2} 2 \theta}{3}\right) \ln (\tan \theta)\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \frac{1}{2}\left(\sin 2 \theta-\frac{\sin ^{3} 2 \theta}{3}\right) \cdot \frac{1}{\tan \theta} \cdot \sec ^{2} \theta d\theta \right)\\ =& 4 \int_{0}^{\frac{\pi}{4}}\left(1-\frac{\sin ^{2} 2 \theta}{3}\right) d \theta \\ =& 4\left(\frac{\pi}{4}\right)-\frac{4}{3} \int_{0}^{\frac{\pi}{4}} \frac{1-\cos 4 \theta}{2} d \theta \\ =& \frac{5 \pi}{6} \end{aligned} $$
$ \text{We are going to evaluate the integral}$ $\displaystyle I=\int_{0}^{1} \frac{x^{3} \ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^{2}}} d x. \tag*{}$ $\textrm{using integration by parts only.}$
$\textrm{Noting that}$ $\displaystyle \quad \frac{d}{d x} \ln \left(\frac{1+x}{1-x}\right)=\frac{2}{1-x^{2}},\tag*{}$ $\textrm{we prefer to using integration by parts.}$ $\begin{align*} \displaystyle I&= \quad \int \frac{x^{3}}{\sqrt{1-x^{2}}} d x \\&=\displaystyle \int \frac{x-x\left(1-x^{2}\right)}{\sqrt{1-x^{2}}} d x \\&=\displaystyle \int \frac{x}{\sqrt{1-x^{2}}} d x-\int x \sqrt{1-x^{2}} d x \\&=\displaystyle -\sqrt{1-x^{2}}+\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{3}+c \\\displaystyle &=-\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)+c\end{align*} \tag*{} $ $\textrm{We can now start the evaluation.}$ $ \displaystyle \begin{aligned}I\displaystyle =&\int_{0}^{1} \ln \left(\frac{1+x}{1-x}\right) d\left(-\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\right) \\\displaystyle =&-\left[\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\ln \left(\frac{1+x}{1-x}\right)\right]_{0}^{1}+\int_{0}^{1} \frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\left(\frac{2}{1-x^{2}}\right) d x \\ \displaystyle =&\frac{2}{3} \int_{0}^{1} \frac{2+x^{2}}{\sqrt{1-x^{2}}} d x\end{aligned} \tag*{}$
$\text {As usual, we let } x=\sin \theta ,$ $\displaystyle \begin{aligned} \int_{0}^{1} \frac{2+x^{2}}{\sqrt{1-x^{2}}} d x=&\int_{0}^{\frac{\pi}{2}} \frac{2+\sin ^{2} \theta}{\sqrt{1-\sin ^{2} \theta}} \cdot \cos \theta d \theta \\&=[2 \theta]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \theta}{2} d \theta \\&=\pi+\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \\&=\frac{5 \pi}{4}\end{aligned} \tag*{} $
We can now conclude that $\displaystyle \boxed{\int_{0}^{1} \frac{x^{3} \ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^{2}}} d x =\frac{2}{3}\cdot \frac{5 \pi}{4}=\frac{5 \pi}{6}}\tag*{} $