# Partial fractions...stuck on one part

• Oct 17th 2010, 02:18 PM
janedoe
Partial fractions...stuck on one part
∫ dx/(2x-1)³

Here's what I have so far:

1 = A(2x-1)² + B(2x-1) + C

[solve for c]
x=1/2
1= A(0)² + B(0) + C
1=C

But in order to solve for A or B, I'd have to use x=1/2, but that makes BOTH terms 0.

Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?
• Oct 17th 2010, 02:27 PM
e^(i*pi)
Substitution would be better for this question

Let $u = 2x-1$

$\dfrac{du}{dx} = 2 \: \rightarrow \: dx = \dfrac{1}{2}du$

$\int \dfrac{\frac{1}{2}du}{u^3} = \dfrac{1}{2} \int \dfrac{du}{u^3} = \dfrac{1}{2} \int u^{-3}\, du$
• Oct 17th 2010, 02:33 PM
skeeter
Quote:

Originally Posted by janedoe
∫ dx/(2x-1)³

Here's what I have so far:

1 = A(2x-1)² + B(2x-1) + C

[solve for c]
x=1/2
1= A(0)² + B(0) + C
1=C

But in order to solve for A or B, I'd have to use x=1/2, but that makes BOTH terms 0.

Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?

why partial fractions?

let $u = 2x-1$

$du = 2\, dx$

$\displaystyle \frac{1}{2} \int \frac{2}{(2x-1)^3} \, dx$

$\displaystyle \frac{1}{2} \int \frac{du}{u^3}$

$\displaystyle -\frac{1}{4u^2} + C$

$\displaystyle -\frac{1}{4(2x-1)^2} + C$
• Oct 17th 2010, 04:04 PM
janedoe
wow i totally over-thought that, thanks guys!
• Oct 17th 2010, 04:45 PM
TheCoffeeMachine
Quote:

Originally Posted by janedoe
Here's what I have so far:

1 = A(2x-1)² + B(2x-1) + C

When you set-up the partial fractions as $\displaystyle \frac{1}{(2x-1)^3} = \frac{a}{2x-1}+\frac{b}{(2x-1)^2}+\frac{c}{(2x-1)^3}$,
you can easily see that you will have:

$\displaystyle \int \frac{1}{(2x-1)^3}\;{dx} = \int\left\{\frac{a}{2x-1}+\frac{b}{(2x-1)^2}+\frac{c}{(2x-1)^3}\right\}\;{dx}$

$\displaystyle= a\int\frac{1}{2x-1}\;{dx}+b\int\frac{1}{(2x-1)^2}\;{dx}+c\int\frac{1}{(2x-1)^3}\;{dx}$

Now, the last integral is the one which we have started with (times a constant).
So our fraction was more desirable in its original form.

Quote:

Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?
That's correct. It's an identity so any random number will do. Having found [LaTeX ERROR: Convert failed] , putting $x = 1$
and $x = -1$ you will find that [LaTeX ERROR: Convert failed] , so in full circle you will end-up with:

[LaTeX ERROR: Convert failed]