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Math Help - Integration of Logaritmic Functions

  1. #1
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    Integration of Logaritmic Functions

    <br />
\int_1^3 (x+3)/(x^2+6x) dx = ?<br />

    I tired separating and simplifying the problem into two different integrals:

    <br />
\int_1^3 1/(x+6) + <br />
\int_1^3 3/(x^2+6x)<br />

    and get...
     \int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x)

    evaluating I get

     <br />
[ln9-ln7]+ 3[ln27+ln7]<br />

    but that is not any of the solutions.
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  2. #2
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    Try Integration by Substitution, which is basically the Chain Rule backward:

    \int\,f'(g(x))g'(x)\,dx\,=\,f(g(x))\,+\,C

    Letting u\,=\,g(x)\,=\,x^2\,+\,6x, we can see that du\,=\,g'(x)\,dx\,=\,(2x\,+\,6)\,dx\,=\,2(x\,+\,3)  \,dx, so that the integral becomes

    \int_{u\,=\,7}^{u\,=\,27}\,\frac{1}{2u}\,du.

    Hope this helps.
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  3. #3
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    Natural log is supposed to be in the answer...any ideas?
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  4. #4
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    Hi frog09,

    Do you understand what Scott H says in #2? He has almost completely worked the problem for you.

    Cant you figure out what \int_{u\,=\,7}^{u\,=\,27}\,\frac{1}{2u}\,du is ?


    If you integrate the above you will see the "supposed natural logs"
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  5. #5
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    Thumbs up

    yes! a moment of enlightenment. thanks!
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by frog09 View Post
    <br />
\int_1^3 (x+3)/(x^2+6x) dx = ?<br />

    I tired separating and simplifying the problem into two different integrals:

    <br />
\int_1^3 1/(x+6) + <br />
\int_1^3 3/(x^2+6x)<br />

    and get...
     \int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x)

    evaluating I get

     <br />
[ln9-ln7]+ 3[ln27+ln7]<br />

    but that is not any of the solutions.


    \begin{gathered}\int\limits_1^3 {\frac{{x + 3}}{{x^2  + 6x}}dx}  = \frac{1}{2}\int\limits_1^3 {\frac{{2x + 6}}{{x^2  + 6x}}dx = } \frac{1}{2}\int\limits_1^3 {\frac{{d\left( {x^2  + 6x} \right)}}{{x^2  + 6x}} = }  \hfill \\= \left. {\frac{1}{2}\ln \left( {x^2  + 6x} \right)} \right|_1^3  = \frac{1}{2}\left( {\ln 27 - \ln 7} \right) = \frac{1}{2}\left( {3\ln 3 - \ln 7} \right). \hfill \\\hfill \\ \end{gathered}
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  7. #7
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    Are you sure? I thought we had to change the bounds of integration after the change of variables.

    Frog09, remember also that \ln\,u is defined as \int_1^u\,\frac{1}{x}\,dx.
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  8. #8
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    Except that he didn't actually change the variable, so it was ok. This is how I would keep track of it...



    (As usual straight continuous lines diff or anti-diff with respect to x, straight dashed lines with respect to the dashed balloon expression.)

    Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
    Last edited by tom@ballooncalculus; January 2nd 2009 at 12:55 PM. Reason: typo
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  9. #9
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Scott H View Post
    Are you sure? I thought we had to change the bounds of integration after the change of variables.

    Frog09, remember also that \ln\,u is defined as \int_1^u\,\frac{1}{x}\,dx.
    Yes, I am sure!
    In my solution we must not change the bounds of integration.
    Because

    \left. {\int\limits_b^a {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = } \ln \left( {f\left( x \right)} \right)} \right|_b^a  = \ln \left| {f\left( a \right)} \right| - \ln \left| {f\left( b \right)} \right|.
    Last edited by DeMath; January 3rd 2009 at 07:44 AM.
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  10. #10
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    All right, I see.
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