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How to integrate y = ln(x)

  1. Aug 28, 2005 #1
    this is an honest question. i have attempted this but i didnt know how to do this. tomorrow is my exam and i realised that there was no where of finding the integration of y=lnx. everytime i ask tis question, people will tell be 1/x but that is the derivative, i am after the integral

    so is there an integral of ln x?
  2. jcsd
  3. Aug 28, 2005 #2


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    Have you learned about integrating by parts?
  4. Aug 28, 2005 #3
    no, does that go into maths c?
    this is only a maths b class
    does this mean, there is no simple way of doing this?
  5. Aug 28, 2005 #4
    [tex] \int ln x dx = x \ln x + x [/tex] i think, but im not sober. its done by integration by parts which looks something like this

    [tex] \int u dv = uv - \int v du [/tex]
  6. Aug 28, 2005 #5


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    It should be [itex] \int ln x dx = x \ln x - x = x(\ln x - 1) [/itex]
  7. Aug 28, 2005 #6
    Yeh, int by parts needs two functions so you have ln(x) and 1. Make u=ln(x) and dv=1, then follow whozums equation and you'll get what TD wrote! +c of course
  8. Aug 28, 2005 #7


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    The other way is to make the obvious substitution [itex]x = e^u[/tex], then integrate by parts.
  9. Aug 28, 2005 #8


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    And indeed you can use the same method to find the integral of any function for which the inverse function has a simple integral.

    Let's say you know that [tex]\int f(x)dx = g(x) + c[/tex]. You want to find [tex]\int f^{-1}(x)dx[/tex] where the exponent denotes the inverse function.

    Substitute [tex]x = f(u)[/tex].

    You get [tex]\int f^{-1}(x)dx = \int uf'(u)du = uf(u) - \int f(u)du = uf(u) - g(u) + c = xf^{-1}(x) - g(f^{-1}(x)) + c[/tex]

    You can verify that works by trying out [tex]f^{-1}(x) = \ln x[/tex] in the result.
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