# Thread: Integration by substituion and partial factions

1. ## Integration by substituion and partial factions

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

2. 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

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

3. Here is what I purpose:

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

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

5. Yeah, that's what I got, too.

6. 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.

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.

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

9. $\displaystyle \frac{1}{2\sqrt{x}}\,dx$ 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.

10. Ah that does make sense. Many thanks!