How do I find the integral of ln x?

[lsim16]

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?

 

[Tide]

Have you learned about integrating by parts?

 

[lsim16]

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?

 

[whozum]

\int ln x dx = x \ln x + x i think, but I am not sober. its done by integration by parts which looks something like this

\int u dv = uv - \int v du

 

[TD]

\int ln x dx = x \ln x + x i think, but I am not sober. its done by integration by parts which looks something like this

\int u dv = uv - \int v du

It should be \int ln x dx = x \ln x - x = x(\ln x - 1)

 

[patcho]

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

 

[Curious3141]

The other way is to make the obvious substitution x = e^u[/tex], then integrate by parts.

 

[Curious3141]

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 \int f(x)dx = g(x) + c. You want to find \int f^{-1}(x)dx where the exponent denotes the inverse function.

Substitute x = f(u).

You get \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

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