Partial Fraction Integration problem

[Jimbo57]

Homework Statement

Homework Equations

The Attempt at a Solution

I have to solve this question and I know that partial fractions is the intended method. I can do the long division easy, which gives:

Setting up for A and B, I get:

which produces:

4x-15= A(x-2)² + B(x-2)

From here, I have no idea how to use Gaussian elimination or x values to isolate for A and B. The answers are A=4 and B=-7, but I don't know how they got there. Anyone able to give me a nudge?

 

[SteamKing]

I don't think your last step is correct.

If you want to clear the (x-2)^2 term from the denominator, you multiply both sides of the equation by that factor, which leads to the equation,

4x - 15 = A*(x-2) + B

which, on expanding, gives

4x - 15 = A*x - 2*A + B

equating the coefficients of the various terms,

4x = A*x

which implies A = 4, and -15 = B - 2* A or -15 = B - 2*4, or B = -7

 

[Jimbo57]

I don't think your last step is correct.

If you want to clear the (x-2)^2 term from the denominator, you multiply both sides of the equation by that factor, which leads to the equation,

4x - 15 = A*(x-2) + B

which, on expanding, gives

4x - 15 = A*x - 2*A + B

equating the coefficients of the various terms,

4x = A*x

which implies A = 4, and -15 = B - 2* A or -15 = B - 2*4, or B = -7

I guess I didn't understand as well as I thought! Thanks so much for pointing this out SteamKing!

 

[HallsofIvy]

Since
[tex]\frac{4x- 15}{(x- 2)^2}= \frac{A}{(x- 2)}+ \frac{B}{(x- 2)^2}[/tex]
or, equivalently,
[tex]4x- 15= A(x- 2)+ B[/tex]
is to be true for all x, choosing any two values for x will give two equations to solve for A and B.

Obviously, x= 2 will simplify a lot: 4(2)- 15= -7= B. Taking, say, x= 0 (just because it is simple) gives -15= -2A+ B and we know that B= -7: -15= -2a- 7 so 2a= -7+ 15= 8 and a= 4.

But any two values of x will give two equations for A and B.