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Thread: Evaluating Partial Fractions

  1. #1
    Junior Member
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    Evaluating Partial Fractions

    Hi,
    This question is bugging me.

    Evaluate the Integral

    (x)/(x^2+7x+6) from 0 to 4

    Solving for it I get (6/5)ln (x+6) - (1/5) ln (x+1)

    Plugging values in I get (6/5) ln 10 - (6/5) ln 6 - (1/5) ln 5

    The answer is (-6/5) ln 3 + ln 5.

    How did this come about?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by shogunhd View Post
    Evaluate the Integral

    (x)/(x^2+7x+6) from 0 to 4
    Since I don't use partial fractions technique frequently, you can do this as follows:

    $\displaystyle \int_0^4 {\frac{x}
    {{x^2 + 7x + 6}}\,dx} = \frac{1}
    {5}\int_0^4 {\frac{{6(x + 1) - (x + 6)}}
    {{(x + 1)(x + 6)}}\,dx} .$

    So $\displaystyle \int_0^4 {\frac{x}
    {{x^2 + 7x + 6}}\,dx} = \frac{1}
    {5}\left( {6 \cdot \ln \left| {x + 6} \right|\Big|_0^4 - \ln \left| {x + 1} \right|\Big|_0^4} \right).$

    Hence $\displaystyle \int_0^4 {\frac{x}
    {{x^2 + 7x + 6}}\,dx} = \frac{1}
    {5}\left( {6\ln 10 - 6\ln 6 - \ln 5} \right).$

    Now here's the make-up:

    $\displaystyle \begin{aligned}
    \frac{1}
    {5}\left( {6\ln 10 - 6\ln 6 - \ln 5} \right) &= \frac{1}
    {5}\left( {6\ln 2 + 6\ln 5 - 6\ln 2 - 6\ln 3 - \ln 5} \right)\\
    &= \frac{1}
    {5}\left( {5\ln 5 - 6\ln 3} \right)\\
    &= \ln 5 - \frac{6}
    {5}\ln 3,\end{aligned}$

    which gives the desired answer.
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