<s>[INST] 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?
 [/INST] 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 $x=\sin t \Rightarrow dx=\cos tdt$:
$$\int \frac {dx}{(1-x^2)^{3/2}}=\int \frac {\cos tdt}{(\cos^2t)^{3/2}}=\int \frac{dt}{\cos^2t}=\\
=\tan t+C=\frac{\sin t}{\sqrt{1-\sin^2t}}+C=\frac{x}{\sqrt{1-x^2}}+C.$$
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