Solving Indefinite Integral $\int{\frac{x^2}{\sqrt{1-x}}}dx$

Question

I'm stuck with this indefinite integral;

$\int{\frac{x^2}{\sqrt{1-x}}}dx$

I've tried and failed to solve this by parts;

$\int{u\space dv} = uv- \int{v\space du}$

$u = (1 - x)^{-\frac{1}{2}}$

$du = \frac{1}{2}(1 - x)^{-\frac{3}{2}}dx$

$v = \frac{1}{3}x^3$

$dv = x^2dx$

But I get stuck again after writing out the final integral; $\int{v\space du}$.

Am I approaching this problem in the right way, and if so, where do I go from here?

Answer 1

Hint:

Set $\sqrt{1-x}=u\implies x=1-u^2\implies dx=-2u\ du$

Answer 2

$$\frac{x^2}{\sqrt{1-x}}=\frac{x^2-2x+1+2x-2+1}{\sqrt{1-x}}=(1-x)^{\frac{3}{2}}-2(1-x)^{\frac{1}{2}}+(1-x)^{-\frac{1}{2}}$$

Answer 3

Hint:

Decompose $$\frac{x^2}{\sqrt{1-x}}=\frac{(1-(1-x))^2}{\sqrt{1-x}}=\frac1{\sqrt{1-x}}-2\frac{1-x}{\sqrt{1-x}}+\frac{(1-x)^2}{\sqrt{1-x}}.$$

The rest is easy.

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Alternatively, by setting $u=1-x$, you will get an $u^{-1/2}$ factor, while the rest will be a polynomial in $u$. After distribution, a linear combination of half-integer powers of $u$, which is straightforward.

$$-\int\frac{(1-u)^2}{\sqrt u}du=\cdots$$