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Math Help - Need Help with Complicated Algebra Transformation in an Integral

  1. #1
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    Need Help with Complicated Algebra Transformation in an Integral

    Today in math class my teacher performed a complicated transformation in order to solve an integral. He changed:

    1/[(x-2)(x+2)]

    to (1/4)/(x-2) - (1/4)/(x+2)

    This transformation was necessary to solve the problem. He did not tell me however how he did this and left it up to me to find. Does anyone know how this was done?
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  2. #2
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    Quote Originally Posted by PD1337 View Post
    Today in math class my teacher performed a complicated transformation in order to solve an integral. He changed:

    1/[(x-2)(x+2)]

    to (1/4)/(x-2) - (1/4)/(x+2)

    This transformation was necessary to solve the problem. He did not tell me however how he did this and left it up to me to find. Does anyone know how this was done?
    He used a Partial Fractions decomposition.

    Recall from when you were doing fractions that

    \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}.


    He has had to go from the right hand side to the left hand side.


    We write

    \frac{1}{(x -2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2}

    \frac{1}{(x - 2)(x + 2)} = \frac{A(x + 2) + B(x - 2)}{(x - 2)(x + 2)}.


    Clearly the numerators are equal...

    A(x + 2) + B(x - 2) = 1

    Ax + 2A + Bx - 2B = 1

    (A + B)x + 2A - 2B = 0x + 1.


    Equating like powers of x gives

    A + B = 0 and 2A - 2B = 1

    A = -B


    2(-B) - 2B = 1

    -4B = 1

    B = -\frac{1}{4}.


    A = -B

    A = \frac{1}{4}.



    Therefore, finally

    \frac{1}{(x - 2)(x + 2)} = \frac{\frac{1}{4}}{x - 2} - \frac{\frac{1}{4}}{x + 2}.

    Now it is easy to integrate.
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  3. #3
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    That's great. Thanks so much for your help.
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