# Partial fraction denominator factorization order

• Oct 6th 2011, 12:55 AM
MechMon
Partial fraction denominator factorization order
Considering

$\displaystyle \int \frac{1}{x^2-4} dx$

I notice that the order of the factoring of the denominator changes the result.
Since
$\displaystyle (x+2)(x-2)=(x-2)(x+2)$

if I choose
$\displaystyle 1=A(x+2)+B(x-2)$

$\displaystyle \frac{1}{4} \ln\left(\frac{x+2}{x-2}\right)$

However if I chose
$\displaystyle 1=A(x-2)+B(x+2)$

the solution is
$\displaystyle \frac{1}{4} \ln\left(\frac{x-2}{x+2}\right)$

Which is correct?
• Oct 6th 2011, 01:09 AM
mr fantastic
Re: Partial fraction denominator factorization order
Quote:

Originally Posted by MechMon
Considering

$\displaystyle \int \frac{1}{x^2-4} dx$

I notice that the order of the factoring of the denominator changes the result.
Since
$\displaystyle (x+2)(x-2)=(x-2)(x+2)$

if I choose
$\displaystyle 1=A(x+2)+B(x-2)$

$\displaystyle \frac{1}{4} \ln\left(\frac{x+2}{x-2}\right)$

However if I chose
$\displaystyle 1=A(x-2)+B(x+2)$

the solution is
$\displaystyle \frac{1}{4} \ln\left(\frac{x-2}{x+2}\right)$

Which is correct?

The order is irrelevant. If you're getting different answers it will be because of a mistake somewhere on your part. If you show all your working, that mistake might be spotted. By the way, an obvious error is use of brackets ( ) rather than mod | |.
• Oct 6th 2011, 02:12 AM
MechMon
Re: Partial fraction denominator factorization order
Got it, thank you. For posterity...

$\displaystyle \int \frac{1}{x^2-4}=\int \frac{1}{(x+2)(x-2)}$ case 1
or
$\displaystyle \int \frac{1}{(x-2)(x+2)}$ case 2

so...
Case1:
$\displaystyle \frac{1}{x^2-4}= \frac{A}{(x+2)}+\ \frac{B}{(x-2)} \Longrightarrow 1=A(x-2)+B(x+2)$

$\displaystyle @x=2: B=\frac{1}{4} @x=-2: A=-\frac{1}{4}$
then
$\displaystyle \int \frac{1}{x^2-4}=\int \frac{-1/4}{(x+2)}+\int \frac{1/4}{(x-2)}=\int \frac{1/4}{(x-2)}-\int \frac{1/4}{(x+2)}$ (***)

or Case 2:
$\displaystyle \frac{1}{x^2-4}= \frac{A}{(x-2)}+ \frac{B}{(x+2)} \Longrightarrow 1=A(x+2)+B(x-2)$
$\displaystyle @x=2: B=-\frac{1}{4} @x=-2: A=\frac{1}{4}$
and
$\displaystyle \int \frac{1/4}{(x-2)}+\int \frac{-1/4}{(x+2)}=\int \frac{1/4}{(x-2)}-\int \frac{1/4}{(x+2)}$(***)

Thus either case(***) works and
$\displaystyle \int \frac{1}{x^2-4}=\frac{1}{4} \ln\left(\frac{x-2}{x+2}\right)$
• Oct 6th 2011, 01:38 PM
mr fantastic
Re: Partial fraction denominator factorization order
Quote:

Originally Posted by MechMon
Got it, thank you. For posterity...

$\displaystyle \int \frac{1}{x^2-4}=\int \frac{1}{(x+2)(x-2)}$ case 1
or
$\displaystyle \int \frac{1}{(x-2)(x+2)}$ case 2

so...
Case1:
$\displaystyle \frac{1}{x^2-4}= \frac{A}{(x+2)}+\ \frac{B}{(x-2)} \Longrightarrow 1=A(x-2)+B(x+2)$

$\displaystyle @x=2: B=\frac{1}{4} @x=-2: A=-\frac{1}{4}$
then
$\displaystyle \int \frac{1}{x^2-4}=\int \frac{-1/4}{(x+2)}+\int \frac{1/4}{(x-2)}=\int \frac{1/4}{(x-2)}-\int \frac{1/4}{(x+2)}$ (***)

or Case 2:
$\displaystyle \frac{1}{x^2-4}= \frac{A}{(x-2)}+ \frac{B}{(x+2)} \Longrightarrow 1=A(x+2)+B(x-2)$
$\displaystyle @x=2: B=-\frac{1}{4} @x=-2: A=\frac{1}{4}$
and
$\displaystyle \int \frac{1/4}{(x-2)}+\int \frac{-1/4}{(x+2)}=\int \frac{1/4}{(x-2)}-\int \frac{1/4}{(x+2)}$(***)

Thus either case(***) works and
$\displaystyle \int \frac{1}{x^2-4}=\frac{1}{4} \ln\left(\frac{x-2}{x+2}\right)$

As mentioned in my first reply, | | modulus NOT ( ) brackets are required around what you're logging. Also, the arbitrary constant of integration needs to be included. Omitting either or both of these will result in marks being deducted.