Evaluate the indefinite integral $\int {(1-x^2)^{-3/2}}dx$ Evaluate the indefinite integral

$$\int \frac {dx}{(1-x^2)^{3/2}} .$$

My answer: I have taken $u = 1-x^2$ but I arrived at the integral of $\frac{-1}{(u^3 - u^4)^{1/2}}$. What should I do next?
 A: Before edit: $\int \frac {dx}{(1-x^2)^3/2}$.
You can decompose the fraction as follows, then easily integrate:
$$\frac {2}{(1-x^2)^3}=\frac14\cdot \left[\frac {1}{1-x}+\frac{1}{1+x}\right]^3=\\
 \frac{1}{4(1-x)^3} +\frac{3}{4(1-x)(1+x)}\left[\frac1{1-x}+\frac1{1+x}\right] + \frac{1}{4(1+x)^3}=\\
 \frac{1}{4(1-x)^3} +\frac38\cdot\left[\frac1{1-x}+\frac1{1+x}\right]^2 + \frac{1}{4(1+x)^3}=\\
 \frac{1}{4(1-x)^3} +\frac{3}{8(1-x)^2}+\frac38\cdot \left[\frac1{1-x}+\frac1{1+x}\right] + \frac{3}{8(1+x)^2}+ \frac{1}{4(1+x)^3}.$$
After edit: $\int \frac {dx}{(1-x^2)^{3/2}}$.
Make the change $x=\sin t \Rightarrow dx=\cos tdt$:
$$\int \frac {dx}{(1-x^2)^{3/2}}=\int \frac {\cos tdt}{(\cos^2t)^{3/2}}=\int \frac{dt}{\cos^2t}=\\
=\tan t+C=\frac{\sin t}{\sqrt{1-\sin^2t}}+C=\frac{x}{\sqrt{1-x^2}}+C.$$
A: Hint: Let $x=\sin t$ then the integral is
$$I=\int1+\tan^2t\ dt$$
A: Hint The occurrence of the form $(1 - x^2)^{3 / 2}$ suggests using the substitution $$x = \sin \theta, \qquad dx = \cos \theta \,d\theta$$ or the substitution $$x = \tanh t, \qquad dx = \operatorname{sech}^2 t .$$

 For the latter, carrying out the substitution and using a hyperbolic Pythagorean identity gives $$\int \frac{dx}{(1 - x^2)^{3 / 2}} = \int \frac{\operatorname{sech}^2 t \,dt}{\operatorname{sech}^3 t} = \int \cosh t\,dt = \sinh t + C ,$$ and $$\sinh t = \frac{\tanh t}{\operatorname{sech} t} = \frac{\tanh t}{\sqrt{1 - \tanh^2 t}} = \frac{x}{\sqrt{1 - x^2}},$$ so $$\color{#df0000}{\boxed{\int \frac{dx}{(1 - x^2)^{3 / 2}} = \frac{x}{\sqrt{1 - x^2}} + C}} .$$

A: Use partial fractions decomposition:$$\frac2{(1-x^2)^3}=\frac{3}{8(x+1)}+\frac3{8(x+1)^2}+\frac1{4(x+1)^3}-\frac3{8(x-1)}+\frac3{8(x-1)^2}-\frac1{4 (x-1)^3}.$$
