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Integral of ln

  1. Jul 2, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]f(x)=\int\frac{x+1}{x^2+2x}[/itex]


    3. The attempt at a solution

    I know that this can be solved with substitution. I'm wondering why I get the wrong answer when using this other method...

    [itex]f(x)=\int\frac{x+1}{x^2+2x}[/itex]

    [itex]=\int\frac{x}{x^2+2x}+\frac{1}{x^2+2x}[/itex]


    [itex]=\int\frac{1}{x+2}+\int\frac{1}{x^2+2}[/itex]

    [itex]=ln(x+2)+ln(x^2+2) + c[/itex]

    I could simplify more, but it's wrong. It should be [itex].5(x^2+2) + c[/itex]

    I'm guessing that [itex]\int\frac{1}{x^2+2}\neq ln(x^2+2)[/itex] so in what cases does [itex]\int \frac{1}{x} = ln(x) + c[/itex] not hold?? I just learned this stuff today so any answers will have to be rather explicit please.
     
  2. jcsd
  3. Jul 2, 2012 #2
    This.

    d/dx ln(x2+2) = 2x/(x2+2), which is obviously not equal to 1/(x2+2). Never forget the chain rule.

    To integrate 1/(x2+2), you need to use trigonometric substitution.
     
    Last edited by a moderator: Jul 2, 2012
  4. Jul 2, 2012 #3
    Don't know that yet. So I guess that any time there's a nested function then ∫(1/x) dx=ln(x)+c doesn't hold?
     
    Last edited by a moderator: Jul 2, 2012
  5. Jul 2, 2012 #4
    Do you know of the chain rule?
     
    Last edited by a moderator: Jul 2, 2012
  6. Jul 2, 2012 #5

    SammyS

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    Of course [itex]\displaystyle \int \frac{1}{x}\,dx=\ln|x|+C[/itex] always holds.

    What doesn't hold in general is [itex]\displaystyle \int \frac{1}{f(x)}\,dx=\ln|f(x)|+C[/itex]
     
    Last edited by a moderator: Jul 2, 2012
  7. Jul 2, 2012 #6
    Yes, I meant I don't know trig substitution.
     
    Last edited by a moderator: Jul 2, 2012
  8. Jul 2, 2012 #7
    Could you provide a few examples please?
     
    Last edited by a moderator: Jul 2, 2012
  9. Jul 2, 2012 #8

    Mark44

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    You did this problem the hard way, by splitting in into two integrals. An easy substitution will do the trick.

    Let u = x2 + 2x, so du = (2x + 2)dx
    Notice any similarity to the numerator of your original integral?
     
    Last edited: Jul 2, 2012
  10. Jul 2, 2012 #9

    Dick

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    It's not a trig substitution. It's a u substitution. Try putting u=x^2+2x. What is du?
     
  11. Jul 2, 2012 #10
    To the above two posts -

    I know it's not trig substitution, I only mentioned I don't know that because it was mentioned by someone else. U-sub is easy, but I was trying to solve it this way because I'm trying to get a handle on derivatives and integrals of ln(x)
     
  12. Jul 2, 2012 #11

    Dick

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    If you don't want to do it the easy way using U-sub, then you'll need to do partial fractions on the second integral. Do you know that?
     
  13. Jul 2, 2012 #12
    Or complete the square and use a trig substitution?
     
  14. Jul 2, 2012 #13

    Dick

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    If doing it the hard way isn't enough and you want to do it the REALLY hard way, that should work.
     
  15. Jul 3, 2012 #14

    SammyS

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    For instance, if f(x) = x2, then

    [itex]\displaystyle \int \frac{1}{f(x)}\,dx=\int\frac{1}{x^2}\,dx=\int x^{-2}\,dx=-x^{-1}+C\ne\ln(x^2)+C[/itex]
     
  16. Jul 3, 2012 #15

    Mentallic

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    [tex]\int\frac{1}{x}dx=\ln|x|+c[/tex] only holds because [tex]\frac{d}{dx}\left(\ln|x|+c\right)[/tex][tex]=\frac{1}{x}\cdot \frac{d}{dx}(x)=\frac{1}{x}[/tex] by use of the chain rule, while, in general, [tex]\frac{d}{dx}\left(\ln|f(x)|+c\right)[/tex][tex]=\frac{1}{f(x)}\cdot\frac{d}{dx}f(x)=\frac{f'(x)}{f(x)}[/tex]

    So if you have some f(x) in the denominator that needs integrating, unless the numerator is the derivative of the denomintor (or at least a constant multiplied by the derivative) then you can't take the log.
     
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