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
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
 A: Substitute $u=\sqrt{1-3x}$ thus $\mathrm{d}x=-\dfrac{2\sqrt{1-3x}}{3}\,\mathrm{d}u$
$$I={\displaystyle\int}\dfrac{2\left(u^2-1\right)}{\sqrt{3}u^2\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u$$
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\left(\dfrac{1}{\left(4-u^2\right)^\frac{3}{2}}-\dfrac{1}{u^2\left(4-u^2\right)^\frac{3}{2}}\right)\mathrm{d}u$$
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{1}{\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u-\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{1}{u^2\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u$$
Now perform trigonometric substitution  $u=2\sin v$ to solve other two integrals
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{2\cos\left(v\right)}{\left(4-4\sin^2\left(v\right)\right)^\frac{3}{2}}\,\mathrm{d}v-\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{\cos\left(v\right)}{2\sin^2\left(v\right)\left(4-4\sin^2\left(v\right)\right)^\frac{3}{2}}\,\mathrm{d}v$$
$$I=\dfrac{1}{2\sqrt{3}}{\displaystyle\int}\dfrac{1}{\cos^2\left(v\right)}\,\mathrm{d}v-\dfrac{1}{8\sqrt{3}}{\displaystyle\int}\dfrac{1}{\cos^2\left(v\right)\sin^2\left(v\right)}\,\mathrm{d}v$$ 
can you solve it from here?
A: Hint:
For real calculus, we need $$-1<x<\dfrac13$$
$$\iff-\dfrac23<x+\dfrac13<\dfrac23$$
WLOG $x+\dfrac13=\dfrac23\cos2t$ where $0<2t<\pi,\cos t,\sin t,\sin2t>0$
$\iff3x+1=2\cos2t,3x=2\cos2t-1=4\cos^2t-3$
$3dx=-8\sin t\cos t\ dt$
$1-3x=1-(2\cos2t-1)=2(1-\cos2t)=4\sin^2t\implies(1-3x)^{3/2}=(2\sin t)^3$
$x+1=\dfrac{3x+3}3=\dfrac{2\cos2t-1+3}3=\dfrac{4\cos^2t}3\implies(1+x)^{3/2}=\dfrac{(2\cos t)^3}{3\sqrt3}$
$$I=\int\dfrac{-8\cdot3\sqrt3(4\cos^2t-3)\sin t\cos t\ dt}{3^28^2\sin^3t\cos^3t}dt$$
$$\implies-8\sqrt3I=\int\dfrac{4\cos^2t-3}{\sin^2t\cos^2t}dt=4\int\sec^2t\ dt-12\int\csc^22t\ dt$$
Can you take it from here?
A: 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
A: Since
$$\dfrac x{(1-3x)(1+x)} = \dfrac14\cdot\dfrac1{1-3x}-\dfrac14\cdot\dfrac1{1+x},$$
then
$$I(x)=\int\dfrac{x\,\mathrm dx}{(1-3x)^{^{^3/_2}}(1+x)^{^{^3/_2}}}
=I_1(x)+I_2(x),\tag1$$
where
$$I_1(x)=\dfrac14\int\dfrac{\mathrm dx}{(1-3x)^{^{^3/_2}}\sqrt{1+x}},\quad
I_2(x) = -\dfrac14\int\dfrac{\mathrm dx}{(1+x)^{^{^3/_2}}\sqrt{1-3x}}.$$
1. 
Substitution 
$$t=\dfrac1{\sqrt{1-3x}},\quad \mathrm dt=\dfrac32\dfrac{\mathrm dx}{(1-3x)^{^{^3/_2}}},\quad x=\dfrac{t^2-1}{3t^2},$$
allows to get
$$I_1=\dfrac{\sqrt3}6\int\dfrac{t\,\mathrm dt}{\sqrt{4t^2-1}} 
=\dfrac{\sqrt3}{24} \sqrt{4t^2-1}+\mathrm{const}
=\dfrac{\sqrt3}{24} \sqrt{\dfrac4{1-3x}-1}+\mathrm{const},$$
$$I_1(x) = \dfrac18\sqrt{\dfrac{1+x}{1-3x}}+\mathrm{const}.\tag2$$
2.
Similarly, substituion
$$t=\dfrac1{\sqrt{1+x}},\quad \mathrm dt=-\dfrac12\dfrac{\mathrm dx}{(1+x)^{^{^3/_2}}},\quad x=\dfrac{1-t^2}{t^2},$$
allows to get
$$I_2 = \dfrac12\int\dfrac{t\,\mathrm dt}{\sqrt{4t^2-3}}
= \dfrac18\sqrt{4t^2-3}+\mathrm{const}
= \dfrac18\sqrt{\dfrac4{1+x}-3}+\mathrm{const},$$
$$I_2(x) = \dfrac18\sqrt{\dfrac{1-3x}{1+x}} + \mathrm{const}.\tag3$$
3.
From $(1)-(3)$ should
$$\color{brown}{\mathbf{I(x) = \dfrac18\sqrt{\dfrac{1+x}{1-3x}} + \dfrac18\sqrt{\dfrac{1-3x}{1+x}} + \mathrm{const}}}.$$
