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Math Help - Partial fraction denominator factorization order

  1. #1
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    Partial fraction denominator factorization order

    Considering

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

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

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

    this leads to:
    \frac{1}{4} \ln\left(\frac{x+2}{x-2}\right)

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

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

    Which is correct?
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  2. #2
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    Re: Partial fraction denominator factorization order

    Quote Originally Posted by MechMon View Post
    Considering

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

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

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

    this leads to:
    \frac{1}{4} \ln\left(\frac{x+2}{x-2}\right)

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

    the solution is
    \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 | |.
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  3. #3
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    Re: Partial fraction denominator factorization order

    Got it, thank you. For posterity...

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

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

    @x=2: B=\frac{1}{4} @x=-2: A=-\frac{1}{4}
    then
    \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:
     \frac{1}{x^2-4}= \frac{A}{(x-2)}+ \frac{B}{(x+2)}  \Longrightarrow 1=A(x+2)+B(x-2)
    @x=2: B=-\frac{1}{4} @x=-2: A=\frac{1}{4}
    and
    \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
    \int \frac{1}{x^2-4}=\frac{1}{4} \ln\left(\frac{x-2}{x+2}\right)
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  4. #4
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    Re: Partial fraction denominator factorization order

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

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

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

    @x=2: B=\frac{1}{4} @x=-2: A=-\frac{1}{4}
    then
    \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:
     \frac{1}{x^2-4}= \frac{A}{(x-2)}+ \frac{B}{(x+2)} \Longrightarrow 1=A(x+2)+B(x-2)
    @x=2: B=-\frac{1}{4} @x=-2: A=\frac{1}{4}
    and
    \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
    \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.
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