# How to integrate y = 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 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?

Homework Helper
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 im not sober. its done by integration by parts which looks something like this

$$\int u dv = uv - \int v du$$

Homework Helper
whozum said:
$$\int ln x dx = x \ln x + x$$ i think, but im 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

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

Homework Helper
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.