Substitution would be better for this question
Let
∫ 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?
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.
That's correct. It's an identity so any random number will do. Having found [LaTeX ERROR: Convert failed] , puttingAm 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?
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.