Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral 

$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$

My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$

By using Euler's 3rd substitution $x = \frac{1-y^2}{3+y^2}$ ($y > 0$, 
so $y = \sqrt{\frac{1-3x}{x+1}}$), we have
$(1-3x)(x+1) = \frac{16y^2}{(3+y^2)^2}$, 
$\sqrt{(1-3x)(x+1)} = \frac{4y}{3+y^2}$, 
$\mathrm{d} x = -\frac{8y}{(3+y^2)^2} \mathrm{d}y$, and hence
\begin{align}
I &= \int \Big(\frac{1}{8} - \frac{1}{8y^2}\Big) \mathrm{d} y\\
&= \frac{1}{8}y + \frac{1}{8y} + C \\
&= \frac{1}{8}\sqrt{\frac{1-3x}{x+1}} + \frac{1}{8}\sqrt{\frac{x+1}{1-3x}} + C\\
&= \frac{1-x}{4\sqrt{(1-3x)(x+1)}} + C.
\end{align}
Euler's 3rd substitution, see: https://en.wikipedia.org/wiki/Euler_substitution
