Evaluate the following indefinite integral: ∫ ((x^2) / (sqrt(1 - x))) dx

Evaluate the following indefinite integral:

∫ ((x^2) / (sqrt(1 - x))) dx

Any and all help is much appreciated! Thanks so much in advance!

∫ x^2 dx/√(1 - x)

Let u = √(1 - x)
then u^2 = 1 - x
and x = 1 - u^2
so dx = -2u du

∫ (1 - u^2)^2 (-2u du)/u = -2 ∫ (1 - u^2)^2 du = -2 ∫ (1 - 2u^2 + u^4) du = -2 (u - 2u^3/3 + u^5/5) + C = -2u + 4u^3/3 - 2u^5/5 + C = -2√(1 - x) + 4/3 (1 - x)^(3/2) - 2 (1 - x)^(5/2) + C

We're going to apply integration by parts twice here.

∫ x^2 / √(1 - x) dx

u = x^2
du = 2x dx
dv = 1/√(1 - x)
v = (-2)(1 - x)^(1/2)

Note: Apply algebraic substitution to integrate 1/√(1 - x).

uv - ∫ vdu
(x^2)(-2)(1 - x)^(1/2) - ∫ (2x)(-2)(1 - x)^(1/2) dx
(x^2)(-2)(1 - x)^(3/2) + 4 ∫ x(1 - x)^(1/2) dx

Applying IBP again:
u = x
du = dx
dv = (1 - x)^(1/2)
v = (-2/3)(1 - x)^(3/2)

uv - ∫ vdu
(x^2)(-2)(1 - x)^(1/2) + 4[(x)(-2/3)(1 - x)^(3/2) - ∫ (-2/3)(1 - x)^(3/2) dx]
(x^2)(-2)(1 - x)^(1/2) + 4[(x)(-2/3)(1 - x)^(3/2) + 2/3 ∫(1 - x)^(3/2) dx]
(x^2)(-2)(1 - x)^(1/2) + 4[(x)(-2/3)(1 - x)^(3/2) + (2/3)(-2/5)(1 - x)^(5/2)] + C
(-2x^2)(1 - x)^(1/2) - (8x/3)(1 - x)^(3/2) - (16/15)(1 - x)^(5/2) + C (Answer)

Have a good day.

-put x=sin^2 t
numerator becomes 2sin^5 t *cost dt
denomenator becomes cos t

now we need to integrate 2 sin^ 5 t with respect to t

∫x²/√(1 - x) dx

u = √(1 - x)

1 - u² = x

-2u du = dx

∫(1 - u²)² * (-2) du

∫(-2 + 4u² - 2u^4) du

= -2√(1 - x) + 4/3*(1 - x)^(3/2) - 2/5*(1 - x)^(5/2) + C