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

Question

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

Answer 1

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?

Answer 2

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?

Answer 3

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

Answer 4

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

$$\color{brown}{\mathbf{I(x) = \dfrac18\sqrt{\dfrac{1+x}{1-3x}} + \dfrac18\sqrt{\dfrac{1-3x}{1+x}} + \mathrm{const}}}.$$