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Math Help - Partial Fractions/ Integration

  1. #1
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    Partial Fractions/ Integration

    Good Day,

    I'm faced with a partial fractions and integration problem I've attached. I got the partial fractions part right.But I just can't get the final answer for the integration part.

    Help need. Thanks.
    Attached Thumbnails Attached Thumbnails Partial Fractions/ Integration-q2.jpg  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by dd86 View Post
    Good Day,

    I'm faced with a partial fractions and integration problem I've attached. I got the partial fractions part right.But I just can't get the final answer for the integration part.

    Help need. Thanks.
    For example,

    -\dfrac{1}{4}\displaystyle\int \dfrac{dx}{-1-x}=\dfrac{1}{4}\left[\log |x+1|\right]_{-10}^{-8}=\dfrac{1}{4}(\log 7-\log 9)=\dfrac{1}{4}\log \dfrac{7}{9}\quad[1]

    \dfrac{1}{4}\displaystyle\int \dfrac{dx}{1-x}=\ldots =\dfrac{1}{4}\log \dfrac{11}{9}\quad [2]

    Now, [1]+[2] ...

    Fernando Revilla
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  3. #3
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    Thank you.

    I was half asleep when I posted this, so I didn't actually state what I had a problem with exactly.

    When I expressed it in partial fractions, I expressed it as (1/4) / (x+1) instead of (-1/4) / (-1-x).

    However, when evaluating it in the integration part, I first left that part out as a log or ln for a negative number does not exist.

    Then, after I saw the answer, I tried by taking the -1 out of (x+1). In the end, I ended up with (1/4) ln (81/77).

    Now I realise that I left out one tiny thing that messed up my calculations .... the modulus sign
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