1. ## Definite integral substitution???

Hi all,

I am struggling to find a way to substitution for integral of x^2/sqr(1-x)dx.
If it was x^3 in the square root I would know how to do it, but because numerator is lower power than denominator

Thanks
alex

2. Originally Posted by rexexdesign
Hi all,

I am struggling to find a way to substitution for integral of x^2/sqr(1-x)dx.
If it was x^3 in the square root I would know how to do it, but because numerator is lower power than denominator

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

let $\displaystyle u=\sqrt{1-x}\Rightarrow{x=1-u^2}$

So $\displaystyle dx=-2udu$

and $\displaystyle x^2=(1-u^2)^2$

So now we have

$\displaystyle \int\frac{-2u(1-u^2)^2}{u}du$

I think you can take it from there

3. You are right. My teacher did not teach us that before. She always said find the outside and inside function and make substitutions, like $\displaystyle \sqrt{x-1}\$ would be $\displaystyle f(x)=\sqrt{x}\$ and $\displaystyle g(x)=x-1$

thanks for the tip.

4. Originally Posted by rexexdesign
You are right. My teacher did not teach us that before. She always said find the outside and inside function and make substitutions, like $\displaystyle \sqrt{x-1}\$ would be $\displaystyle f(x)=\sqrt{x}\$ and $\displaystyle g(x)=x-1$

thanks for the tip.
It is an art form though. As I am sure most will attest on this website it takes a while to see whether you should for this example

Let $\displaystyle u=\sqrt{1-x}$

or Let $\displaystyle u=1-x$