Trig substitution or part integration How can I integrate the following function:
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx $$
? I've tried integration by parts but I think I'm doing in the wrong way.
I need some help!

$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx $$
Substitute $x=\sin(u)$ and $dx=\cos(u) du$. Then $(1-x^2)^{3/2}=(1-\sin^2(u))^{3/2}=\cos^3(u)$ and $u=\sin^{-1}(x)$
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx =$$
$$\int \cot^4(u)\csc^2(u) du$$
For the intergrand $\cot^4(u)\csc^2(u)$ substitute $s=\cot(u)$ and $ds=-\csc^2(u) du$
$$\int \cot^4(u)\csc^2(u) du=\int s^4 ds$$
The integral of $s^4$ is $\frac{s^5}{5}$
$$\int s^4 ds=-\frac{s^5}{5}+C$$
Substitute back for $s=\cot(u)$ and $u=\sin^{-1}(x)$
$$-\frac{s^5}{5}+C=-\frac{1}{5}\cot\left(\sin^{-1}x\right)^5+C$$
Simplify using $\cot(\sin^{-1}(z))=\frac{\sqrt{1-z^2}}{z}$
$$-\frac{1}{5}\cot\left(\sin^{-1}x\right)^5+C=\frac{\left(1-x^2\right)^{5/2}}
{5x^5}+C$$
So:
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx=\frac{\left(1-x^2\right)^{5/2}}
{5x^5}+C$$
