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Homework Help: Integration by substitution,

  1. Jun 13, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \int \frac{3x}{2x+3} [/tex]

    [tex] u = 2x +3 [/tex]

    [tex] x = \frac{1}{2}(u-3} )[/tex]

    [tex] dx = \frac{1}{2} du [/tex]

    so now the integral should be,

    [tex] \int \frac{ \frac{3u-9}{2}}{u} \times \frac{1}{2} du [/tex]

    = [tex] \frac{1}{2} \int \frac{3u-9}{2} \times \frac{1}{u} du [/tex]

    [tex] \frac{1}{2} \int \frac{3u-9}{2u} du [/tex]

    [tex] \frac{1}{2} \int \frac{3u}{2u} - \frac{9}{2u} [/tex]

    [tex] \frac{1}{2} \int \frac{3}{2} - \frac{9}{2u} [/tex]

    = [tex] \frac{1}{2} [ \frac{3}{2}u - \frac{9}{2} ln(2u)] [/tex]

    so final answer, after simplifying is ; [tex] \frac{3}{2}x +\frac{9}{4} - \frac{9}{4}ln(4x+6) +c [/tex]

    but my book says the correct answer is [tex] \frac{3}{2}x - \frac{9}{4}ln(2x+3) +c [/tex]

    So if anyone can tell me where I have gone wrong, I would really appreciate it.

    thanks
     
  2. jcsd
  3. Jun 13, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    The two answers are equivalent, just with different constants of integration. Remember, [itex]\ln(4x+6)=\ln(2)+\ln(2x+3)[/itex], so the additional terms are just constants which can be absorbed into the integration constant.
     
  4. Jun 13, 2010 #3

    Oh okay thanks, I didn't think of it like that, are you also referring to the extra 9/4 term that I get?

    so both answers are valid?
     
  5. Jun 13, 2010 #4
    Yes, so the difference between the two is a constant equal to 9/4 + ln(2).
     
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