Thread: 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?

Substitution would be better for this question

Let

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

wow i totally over-thought that, thanks guys!

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 ,
you can easily see that you will have:





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.

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
and you will find that [LaTeX ERROR: Convert failed] , so in full circle you will end-up with:

[LaTeX ERROR: Convert failed]

which is your original integral.

Morale: don't 'partial fractionalise' a partial fraction.