integral of x^2/√1-x dx?

Put u =√(1 - x)

Then, u² = 1 - x
x = 1 - u²
x² = (1 - u)²
x² = u⁴ - 2u² + 1

Also, x = 1 - u²
dx = -2u du


∫[x² / √(1 - x)] dx
= ∫[(u⁴ - 2u² + 1) / u] (-2u du)
= ∫[-2u⁴ + 4u² - 2] du
= -(2/5)u⁵ + (4/3)u³ - 2u + C
= -(2/5)[√(1 - x)]⁵ - (4/3)[√(1 - x)]³ + 2[√(1 - x)] + C
= -(2/5)[√(1 - x)]⁴√(1 - x) - (4/3)(1 - x)√(1 - x) + 2[√(1 - x)] + C
= -(2/5)(x² - 2x + 1)√(1 - x) - (4/3)(1 - x)√(1 - x) + 2√(1 - x) + C
= -(2/15)[3(x² - 2x + 1) - 10(1 - x) + 15]√(1 - x) + C
= -(2/15)(3x² - 6x + 3 - 10 + 10x + 15)√(1 - x) + C
= -(2/15)(3x² + 4x + 8)√(1 - x) + C

this looks like an integration by parts (I assume you mean square root of (1-x).
rewrite as integral of x^2(1-x)^-1/2 dx.
use formula ∫udv=uv-∫vdu.
1 Choose x^2 as your "u" (always choose a "u" that will get more simple as you take its derivative)
2. choose the rest, (1-x)^-1/2dx as your "dv".
3. find du:
du=2xdx (derivative of x^2 times dx).
4. Find v:
is integral of dv, ∫(1-x)^-½ or 2(1-x)^½.
5. rewrite equation now using formula: 2x^2(1-x)^½ - ∫2(1-x)^1/2*2xdx
since we still cant integrate need to repeat the steps one through five for ∫2(1-x)^1/2*2xdx.

I got 2x^2(1-x)^½ -8/3x(1-x)^3/2+16/15(1-x)^5/2+C. you can factor stuff out but our teacher would be fine with this answer. let me know if you want to see the repeated steps, or there's actually a shortcut to this process. forgive my not pretty notation, this is my first math answer online.

Make the substitution u=sqrt(1-x), du=-dx/[2sqrt(1-x)] and the integral becomes
that of -2(1-u^2)^2 du which is now easy.
Remember to switch the answer back to x
and add C since the integral is indefinite.