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Integral (Partial-Frac Decomp) SIMPLE?

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Integral (Partial-Frac Decomp) **SIMPLE?

Homework Statement


[itex]\int^3_2 \frac{-dx}{x^2-1}[/itex]


Homework Equations





The Attempt at a Solution


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

[itex] -1 = A(x-1) + B(x+0) [/itex]

[itex] -1 = Ax - A + Bx [/itex]

Split into two equations...

[itex] (1):: -1 = -A [/itex]
[itex] (2):: A = 1 [/itex]

[itex] (3):: 0 = A + B [/itex]

Sub (2) -> (3)

[itex] (3):: 0 = 1 + B [/itex]
[itex] (4):: B = -1 [/itex]

So, from above..

[itex] \int^3_2 \frac{-dx}{x^2-1} = \int^3_2 \frac{1}{x} - \frac{1}{x-1} dx [/itex]

[itex] = ln|x| \right|^3_2 - ln|x-1| \right|^3_2 [/itex]

[itex] = ln(3) - ln(2) - ln(2) + ln(1) [/itex]

[itex] = ln(3) - 2ln(2) + ln(1) [/itex]

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
 

Answers and Replies

  • #2


Homework Statement


[tex]
\int^3_2 \frac{-dx}{x^2-1}
[/tex]


[tex]
\int^3_2 \frac{A}{x} + \frac{B}{x-1} dx
[/tex] For some integers A and B.
Can you justify this? I don't think it separates like that.
 
Last edited:
  • #3
LCKurtz
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Homework Statement


[itex]\int^3_2 \frac{-dx}{x^2-1}[/itex]

The Attempt at a Solution


[itex]= \int^3_2 \frac{A}{x} + \frac{B}{x-1} dx[/itex] For some integers A and B.
Wrong at the first step. x2-1 = (x+1)(x-1), not x(x-1).
 

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