Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

∫^[infinity]_2 1/x (ln x)^2 dx

Answer 1

It diverges.

Explanation:

We are given the inetegral:
`\int\limits^(\infty)_2 (1)/(x) (\ln x)^2 dx`

`\int\limits^(\infty)_2 (1)/(x) (\ln x)^2 dx=\int\limits^(\infty)_2 (\ln x)^2 d(\ln x)=\\\\=\lim_(t \to \infty) \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_(t \to \infty) ((\ln t)^3)/(3) |^t_2=\infty-((\ln 2)^3)/(3) =\infty`

So it is divergent.