What is the integral of (x^2/(1-x))dx?

Answer 1

∫ ( x² / (1 - x) dx )

Since the degree of the numerator is greater than the degree of the denominator, synthetic long division would turn this into something more integrable. However, we can also use substitution (and I'm going to, considering how hard it is to show synthetic long division on here.)

Let u = 1 - x. Then
x = 1 - u, and
dx = -du

Our substitution then becomes

∫ [ (1 - u)² / u (-du) ]

Expand the squared binomial.

∫ ( [ 1 - 2u + u² ] / u (-du) )

Factor out the negative next to the du.

(-1) ∫ ( [ 1 - 2u + u² ] / u du )

Divide each term by u.

(-1) ∫ ( [ 1/u - 2 + u ] du )

Integrate term by term.

(-1) ( ln|u| - 2u + (1/2)u²) + C

Distribute the (-1)

-ln|u| + 2u - (1/2)u² + C

But u = 1 - x, so our final answer is

-ln|1 - x| + 2(1 - x) - (1/2)(1 - x)² + C

Answer 2

Perform long division on the rational expression and you get x^2/(1-x) = -x-1 + (1/(1-x)). So the integral is -(x^2)/2 - x- ln(1-x) + C.