Integration by substituion and partial factions

**flashylightsmeow** · April 4th 2011, 05:20 AM

Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).

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

so i substitute $u=\sqrt{x}$ so dx=2u du and form the new integral:

$\int\frac{2u}{2u^2(1-u)(2-u)}\ du$ but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S

Please help!

**mathaddict** · April 4th 2011, 05:35 AM

Originally Posted by flashylightsmeow

Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).

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

so i substitute $u=\sqrt{x}$ so dx=2u du and form the new integral:

$\int\frac{2u}{2u^2(1-u)(2-u)}\ du$ but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S

Please help!

After simplification, you are left with this $2\int \frac{1}{(1-u)(2-u)}du$ , try breaking 1/(1-u)(2-u) again.

**tttcomrader** · April 4th 2011, 05:38 AM

Here is what I purpose:

Reduce your integral to $\int \frac {1}{u(u-1)(u-2)}$ , then perform partial fractions.

**flashylightsmeow** · April 4th 2011, 05:39 AM

Ah, ok. But one thing, doesnt the simplication leave the denominator as u(1-u)(2-u)?

**tttcomrader** · April 4th 2011, 05:41 AM

Yeah, that's what I got, too.

**HallsofIvy** · April 4th 2011, 05:42 AM

Yes, mathaddict miswrote. The u in the numerator cancels one u in the denominator. Now you need A, B, C, so that
$\frac{2}{u(u-1)(u-2)}= \frac{A}{u}+ \frac{B}{u-1}+ \frac{C}u-2}$ . That should be easy.

**Prove It** · April 4th 2011, 05:42 AM

Originally Posted by tttcomrader

Here is what I purpose:

Reduce your integral to $\int \frac {1}{u(u-1)(u-2)}$ , then perform partial fractions.

This does not work.

Notice that if you let $\displaystyle u = \sqrt{x}$ then $\displaystyle du = \frac{1}{2\sqrt{x}}\,dx$ .

So $\displaystyle \int{\frac{1}{\sqrt{x}(1 - \sqrt{x})(2 - \sqrt{x})}\,dx} = 2\int{\left[\frac{1}{(1-\sqrt{x})(2 - \sqrt{x})}\right]\frac{1}{2\sqrt{x}}\,dx}$

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

which can then be solved using partial fractions. Mathaddict is correct.

**flashylightsmeow** · April 4th 2011, 05:49 AM

So the $\sqrt{x}$ that is part of dx doesn't form part of the new integral!?

**Prove It** · April 4th 2011, 05:52 AM

$\displaystyle \frac{1}{2\sqrt{x}}\,dx$ is replaced with $\displaystyle du$ .

Alternatively, $\displaystyle \frac{1}{2\sqrt{x}}$ is replaced with $\displaystyle \frac{du}{dx}$ , so that $\displaystyle \frac{1}{2\sqrt{x}}\,dx = \frac{du}{dx}\,dx = du$ .

But yes, the derivative that you find does not become part of the resulting $\displaystyle u$ integral.

**flashylightsmeow** · April 4th 2011, 05:57 AM

Ah that does make sense. Many thanks!

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