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Math Help - Evaluating this integral

  1. #1
    Junior Member
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    Evaluating this integral

    The integral of 1/(x-3)(x+2)
    The answer is 1/5 ln |x-2/x| + C

    I split it up to be A/(x-3) + B/(x+2) = (Ax+2A+Bx-3B)/(x-3)(x+2) = (x(A+B)+2A-3B)/(x-3)(x+2). Then I set A+B = 0 so 2A-3B must equal 1? Am I doing this right? I don't exactly know where to start by solving for A or B.

    Thank you!
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  2. #2
    Newbie kbzoon's Avatar
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    Note that: <br />
\frac{1}{\left( x-3 \right)\left( x-2 \right)}=\frac{x-2-\left( x-3 \right)}{\left( x-3 \right)\left( x-2 \right)}=\frac{1}{\left( x-3 \right)}-\frac{1}{\left( x-2 \right)}<br />

    <br />
\Rightarrow \int{\frac{dx}{\left( x-3 \right)\left( x-2 \right)}}=\int{\frac{dx}{\left( x-3 \right)}}-\int{\frac{dx}{\left( x-2 \right)}}=\ln \left| x-3 \right|-\ln \left| x-2 \right|+C<br />

    <br />
\int{\frac{dx}{\left( x-3 \right)\left( x-2 \right)}}=\ln \left| \frac{x-3}{x-2} \right|+C<br />

    Another way to get a result.
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  3. #3
    Member Nacho's Avatar
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    Quote Originally Posted by juicysharpie View Post
    I set A+B = 0 so 2A-3B must equal 1? Am I doing this right?
    Thatīs right

    And you can begin solving for A or B, is same
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