How to integrate 1/(1-x^2)^3/2

∫1/(1-x^2)^3/2dx

I tried usub but that doesn't work. I also don't think partial fractions work due to the ^3/2.

We'll use a trig substitution:
x = sin(u)
dx = cos(u) du

∫ 1/(1−x²)^(3/2) dx = ∫ 1/(1−sin²u)^(3/2) * cos(u) du
. . . . . . . . . . . . . . . = ∫ 1/(cos²u)^(3/2) * cos(u) du
. . . . . . . . . . . . . . . = ∫ 1/cos³u * cos(u) du
. . . . . . . . . . . . . . . = ∫ 1/cos²u du
. . . . . . . . . . . . . . . = ∫ sec²u du
. . . . . . . . . . . . . . . = tan(u) + C
. . . . . . . . . . . . . . . = sin(u)/cos(u) + C

Substitute back:
sin(u) = x
cos(u) = √(1 − sin²u) = √(1 − x²)

∫ 1/(1−x²)^(3/2) dx = x/√(1−x²) + C

∫dx/(1 - x²)^(3/2)

Substitute x = sin(t) where t = theta, then integral becomes:

∫cos(t) dt/(cos^3(t)) = ∫sec^2(t) dt = tan(t) + C

= x/√(1 - x²) + C

x=sin u
dx=cos u du
∫1/(1-x^2)^3/2dx
=∫cos u du / cos^3u
=∫sec^2u du
=tan u +C
=sin u/cos u +C
=x/(1-x^2)^1/2 +C