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Integrating (5x)/(3x^2+5) dx

  1. Nov 2, 2012 #1
    Hi all. I am having Calculus 1 this year. We are using a book called Thomas Calculus.
    I think its a lot of fun, but I have to work very much since there is basic stuff like trigonometry that I know really bad. Since I work so much with math I thought it could be fun and helpful to talk with other math people in here:-)

    To the question:

    ∫ (5x)/(3x^2+5) dx

    What I did was this but I think its too complicated:

    ∫ (1/(3x^2+5)) * 5x

    Integration by parts give.

    (5x) ln(3x^2+5) - 5 ∫ ln (3x^2+5)dx

    Not sure have to calculate the last integral. Not sure about anything here really..
     
  2. jcsd
  3. Nov 2, 2012 #2
    compute by substitution:u=3x^2+5
     
  4. Nov 3, 2012 #3
    A new question: With u=3x^2+5 the answer is 5/6*log(3*x^2+5), buit using symbolics in matlab the answer is 5/6*log(x^2+5/3) and that's using u=x^2+5/3. Why are there different answers?
    As 5/6*log(3*x^2+5) = 5/6*log(x^2+5/3)/log(3) <> 5/6*log(x^2+5/3)
     
  5. Nov 3, 2012 #4

    SammyS

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    The two answers differ only by a constant. Remember the constant of integration?

    [itex]\displaystyle \frac{5}{6}\log(3x^2+5)=\frac{5}{6}\log\left(3 \left(x^2+\frac{5}{3}\right)\right)[/itex]
    [itex]\displaystyle =\frac{5}{6}\log(3)+\frac{5}{6}\log\left(x^2+\frac{5}{3}\right)[/itex]​
     
  6. Nov 3, 2012 #5
    Yes of course. I should have thought of that.
    Would you get the same answer using integration by parts?

    If the exercise looked like this instead:

    ∫ (5x)/(3x^2+4x+5) dx

    then you have to use integration by parts?
     
  7. Nov 3, 2012 #6

    tiny-tim

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    welcome to pf!

    hi beaf123! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    integrating by parts, your u would be the whole thing, and your v would be 1 :wink:

    (your line starting "(5x) ln(3x^2+5) …" was wrong)
    no, write the integrand A(6x+4)/(3x2+4x+5) + B/(3x2+4x+5), and do two different substitutions :smile:
     
  8. Nov 6, 2012 #7
    Thank you:)
     
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