Improper Integral #3

[Silvanoshei]

How to setup:

\(\displaystyle y=\frac{2}{4x^{2}-1}?\)

\(\displaystyle \frac{A}{(2x+1)}+\frac{B}{(2x-1)}?\)

If so... how do I find A and B? x-1 and x+2 would make this so much easier.

\(\displaystyle 1=A(2x-1)+B(2x+1)?\)

 

[HallsofIvy]

You want to find two values, A and B, so you need two equations. There are many different ways to do that but you do have to do something and not just try to "remember" a formula.

If you take x= 0, 1= A(2x+ 1)+ B(2x- 1) becomes A- B= 1. If you take x= 1, it becomes 3A+ B= 1. Can you solve those two equations for A and B?

Or if you take x= -1, it becomes -A- 3B= 1. Can you solve the pair of equations A- B= 1, -A- 3B= 1? Or 3A+ B= 1, -A- 3B= 1?

If you take x= 1/2, 1= A(2x+ 1)+ B(2x- 1) become 2A= 1 and if you take x= -1/2, it becomes -2B= 1. Do you see why you might have seen that from the start?