Evaluate the indefinite integral.?

Question

The integral of x^2 / sqrt(1-x) dx. Please show your work. Thank you for your help:)

Answer 1

∫udv = uv - ∫vdu
∫x^2dx/√(1-x)
let u = x^2, dv = dx/√(1-x)
du = 2xdx, v = -2√(1-x)
∫x^2dx/√(1-x) = -2x^2√(1-x) + ∫(4x√(1-x))dx
u = 4x, dv = √(1-x))dx
du = 4dx, v = -(2/3)(1-x)^(3/2)
∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) + ∫(8/3)(1-x)^(3/2)dx
∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) - (16/15)(1-x)^(5/2)
∫x^2dx/√(1-x) = √(1-x)(2x^2 - (8x/3)(1 - x) - (16/15)(1-x)^2)
∫x^2dx/√(1-x) = √(1-x)(2x^2 - 8x/3 - 8x^2/3 - (16/15)(1 - 2x + x^2)
∫x^2dx/√(1-x) = (1/15)√(1-x)(30x^2 - 40x - 40x^2 - 16 + 32x + 16x^2)
∫x^2dx/√(1-x) = -2√(1 - x)(x(3x + 4) + 8)/15

Answer 2

Just transform the variable from x to t^2=(1-x). If you cant figure out how to solve tthe integral even after that, then you really need to take math a bit seriously.