# Math Help - Integration by Substitution

1. ## Integration by Substitution

Hey everyone. I am having trouble with this problem:

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

I can only use substitution to do this. I have no idea what to do. I tried rewriting it in to something that looks integrable using substitution but everything has failed. Here are some of my attempts at rewriting it:

$\sqrt{\frac{x^4}{1-x}}$

$\frac{x^2\sqrt{1+x}}{\sqrt{1-x^2}}$

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

$\frac{\sqrt{x^4-x^5}}{1-x}$

And non of them are integrable by substitution in my eyes. Can someone help me out?

2. Let $y = 1-x$, then you have $\int\frac{x^2}{\sqrt{1-x}}\;{dx} = -\int\frac{(1-y)^2}{\sqrt{y}}\;{dy}$. From there it's
just routine: expand the numerator, divide each term by $\sqrt{y}$ then integrate.

3. Hello, lilaziz1!

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

"Substitution" usually means to let $\,u$ equal something already in the problem.

Let $u\,=\,\sqrt{1-x} \quad\Rightarrow\quad u^2 \,=\,1-x$

Then: . $x \,=\,1-u^2 \quad\Rightarrow\quad dx \,=\,-2u\,du$

Substitute: . $\displaystyle \int\frac{(1-u^2)^2}{u}\,(-2u\,du) \;=\;-2\int(1-u^2)^2\,du$

. . $\displaystyle =\;-2\int \left(1-2u^2+u^4\right)\,du \;=\;-2\left(u - \tfrac{2}{3}u^3 +\tfrac{1}{5}u^5\right) + C$

. . $=\;-\frac{2}{15}u(15 - 10u^2 + 3u^4) + C$

Back-substitute: . $-\frac{2}{15}\sqrt{1-x}\bigg[15 - 10(1-x) + 3(1-x)^2\bigg] + C$

. . $=\;-\frac{2}{15}\sqrt{1-x}\bigg[15 - 10 + 10x + 3 - 6x + 3x^2\bigg] + C$

. . $=\;-\frac{2}{15}\sqrt{1-x}\,(3x^2 + 4x + 8) + C$