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Math Help - 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:

    \int_0^4 {\frac{x}<br />
{{x^2 + 7x + 6}}\,dx} = \frac{1}<br />
{5}\int_0^4 {\frac{{6(x + 1) - (x + 6)}}<br />
{{(x + 1)(x + 6)}}\,dx} .

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

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

    Now here's the make-up:

    \begin{aligned}<br />
\frac{1}<br />
{5}\left( {6\ln 10 - 6\ln 6 - \ln 5} \right) &= \frac{1}<br />
{5}\left( {6\ln 2 + 6\ln 5 - 6\ln 2 - 6\ln 3 - \ln 5} \right)\\<br />
&= \frac{1}<br />
{5}\left( {5\ln 5 - 6\ln 3} \right)\\<br />
&= \ln 5 - \frac{6}<br />
{5}\ln 3,\end{aligned}

    which gives the desired answer.
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