Definite integral substitution???

**rexexdesign** · May 18th 2008, 07:34 PM

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

**Mathstud28** · May 18th 2008, 07:42 PM

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

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

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

So $dx=-2udu$

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

So now we have

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

I think you can take it from there

**rexexdesign** · May 18th 2008, 08:23 PM

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

thanks for the tip.

**Mathstud28** · May 18th 2008, 08:33 PM

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 $\sqrt{x-1}\$ would be $f(x)=\sqrt{x}\$ and $g(x)=x-1$

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 $u=\sqrt{1-x}$

or Let $u=1-x$

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