# Integral (Partial-Frac Decomp) SIMPLE?

• theRukus

#### theRukus

Integral (Partial-Frac Decomp) **SIMPLE?

## 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.

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

$-1 = Ax - A + Bx$

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

theRukus said:

## Homework Statement

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

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

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

Last edited:

theRukus said:

## 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).