Integrate $\int\frac{1}{x^6} \sqrt{(1-x^2)^3} ~ dx$ How to integrate the following? 

$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$


Integral,
$$
\begin{align*}
I &= \int \frac{\sqrt{(1-x^2)^{3}}}{x^6} dx 
\\ &= \int \left( \frac{1-x^2}{x^2} \right)^{3/2} \frac{1}{x^3}dx
\\ &= \int \left(\frac{1}{x^2}-1 \right)^{3/2} \cdot \frac{1}{x^3} dx
\end{align*}
$$
Make the substitution: $z= \frac{1}{x^2}-1$, so that $dz=\frac{-2}{x^3} ~dx$.
$$
\begin{align*}
\text{Integral}
&= -\frac{1}{2}\int z^{3/2} dz 
\\ &= -\frac{1}{2} \cdot \frac{2}{5}z^{5/2}+C
\\ &= -\frac{1}{5} \left(\frac{1}{x^2} - 1 \right)^{5/2}+C
\\ &= - \frac{1}{5} \left(\frac{\sqrt{1-x^2}}{x} \right)^5+C.
\end{align*}
$$
