How do you integrate (1 / y ln y) dy?

Question

or it can be rewritten as (y ln y)^-1 dy

Answer 1

Integrate using the method of substitution based on the fact that the derivative of log(y) + C = 1/y

let u = log(y) then du/dy = 1/y. So du = (1/y) dy

Using these substitutions we have:

Int 1/y*log(y) dy = Int 1/log(y)*(1/y) dy

= Int 1/u du = log(u)+C = log(logy)+C, since u = log(y)

This is correct since from the chain rule:

d/dy log(logy)+C= 1/log(y)*d(logy)/dy= 1/(y*logy)

Answer 2

You cannot find the [ Int (1/(y ln y) ] using the traditional methods of integration that because both of (1/y) and (1/ln y) have infinite number of derivatives.

From the other hand the [ Int { (1/y) ln y} ] equals to ln (ln y) as the guys do it above.

Answer 3

in the name of allah ,the merciful,the gracious.
peace up on u

ok ,im still student but i will try


the differ of ln y is 1/y=y^- 1 ,so i remmber law said

f`(x)*f(x)^n=f(x)^n+1/(n+1)

(y)^-1*(ln y)^-1 dy=(ln y)^-1+1/-1+1 it is not defined
i dont no it is true or not but it just try

note: (^ )mean power

Answer 4

Try substituting u = ln y.

I think you end up with ln(ln y) as an answer.

Answer 5

joe_ska thinks that is (1 / y )*ln y
but it 's int(1/(yLny)) , use maple with this code:
int(1/(y*ln(y)));

Answer 6

Put lny=z
Thus[dy/y]=dz
So now u have integral[ dz/z]=lnz=ln[lny]