Integral (Partial-Frac Decomp) SIMPLE?

[theRukus]

Integral (Partial-Frac Decomp) **SIMPLE?

Homework Statement

\int^3_2 \frac{-dx}{x^2-1}

Homework Equations

The Attempt at a Solution

= \int^3_2 \frac{A}{x} + \frac{B}{x-1} dx For some integers A and B.

-1 = A(x-1) + B(x+0)

-1 = Ax - A + Bx

Split into two equations...

(1):: -1 = -A
(2):: A = 1

(3):: 0 = A + B

Sub (2) -> (3)

(3):: 0 = 1 + B
(4):: B = -1

So, from above..

\int^3_2 \frac{-dx}{x^2-1} = \int^3_2 \frac{1}{x} - \frac{1}{x-1} dx

= ln|x| \right|^3_2 - ln|x-1| \right|^3_2

= ln(3) - ln(2) - ln(2) + ln(1)

= ln(3) - 2ln(2) + ln(1)

I entered the final line into the answer box for my course assignment and it said wrong... So I must be doing something wrong.. =(

Any help is greatly appreciated. Love you PhysicsForums xx

 

[DivisionByZro]

Homework Statement

<br /> \int^3_2 \frac{-dx}{x^2-1}<br /><br /> \int^3_2 \frac{A}{x} + \frac{B}{x-1} dx<br /> For some integers A and B.

Can you justify this? I don't think it separates like that.

 

[LCKurtz]

Homework Statement

\int^3_2 \frac{-dx}{x^2-1}

The Attempt at a Solution

= \int^3_2 \frac{A}{x} + \frac{B}{x-1} dx For some integers A and B.

Wrong at the first step. x²-1 = (x+1)(x-1), not x(x-1).