How do I find the integral of ln x?

  • Thread starter lsim16
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In summary, there are two ways to find the integral of ln x: by using integration by parts or by making the substitution x = e^u and using the inverse function method. Both methods involve breaking down the function into smaller parts and using known integrals to find the solution.
  • #1
lsim16
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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 realized 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?
 
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  • #2
Have you learned about integrating by parts?
 
  • #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?
 
  • #4
[tex] \int ln x dx = x \ln x + x [/tex] i think, but I am not sober. its done by integration by parts which looks something like this

[tex] \int u dv = uv - \int v du [/tex]
 
  • #5
whozum said:
[tex] \int ln x dx = x \ln x + x [/tex] i think, but I am 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
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
The other way is to make the obvious substitution [itex]x = e^u[/tex], then integrate by parts.
 
  • #8
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.
 

1. How do I integrate y = ln(x)?

To integrate y = ln(x), you can use the integration by parts method or the substitution method. Alternatively, you can use the fact that ln(x) is the inverse of e^x and use the inverse function rule to integrate.

2. What is the general formula for integrating ln(x)?

The general formula for integrating ln(x) is ∫ln(x) dx = xln(x) - x + C, where C is the constant of integration.

3. Can I use u-substitution to integrate ln(x)?

Yes, you can use u-substitution to integrate ln(x). Let u = ln(x), then du/dx = 1/x and dx = x du. Substituting into the integral, we get ∫ln(x) dx = ∫u x du = x u - ∫x du = x ln(x) - ∫1 dx = x ln(x) - x + C.

4. Are there any special cases when integrating ln(x)?

Yes, when integrating ln(x), you need to be careful when the limits of integration include 0 or a negative number. In these cases, you will need to use the absolute value of x to avoid undefined values.

5. Can I use integration by parts to integrate ln(x)?

Yes, you can use integration by parts to integrate ln(x). Let u = ln(x) and dv = dx, then du/dx = 1/x and v = x. Substituting into the integration by parts formula, we get ∫ln(x) dx = x ln(x) - ∫x (1/x) dx = x ln(x) - ∫1 dx = x ln(x) - x + C.

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