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How to integrate y = ln(x)

  • Thread starter lsim16
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  • #1
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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?
 

Answers and Replies

  • #2
Tide
Science Advisor
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Have you learned about integrating by parts?
 
  • #3
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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?
 
  • #4
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[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]
 
  • #5
TD
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whozum said:
[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]
It should be [itex] \int ln x dx = x \ln x - x = x(\ln x - 1) [/itex]
 
  • #6
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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
 
  • #7
Curious3141
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The other way is to make the obvious substitution [itex]x = e^u[/tex], then integrate by parts.
 
  • #8
Curious3141
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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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