Question

How do you integrate `(ln(x)*(1/x))dx`?

Answer 1

`= 1/2 (ln x)^2 + C`

Explanation:

`int dx qquad ln(x)*1/x`

you can do this by inspection as `(ln x) ' = 1/x` so we can trial `alpha (ln x)^2` as a solution.

So `(alpha (ln x)^2 + C)' = 2 alpha ln x 1/x implies 2 alpha = 1, alpha = 1/2`

if you don't fancy that you could use IBP : `int uv' = uv - int u'v`

`u = ln x, u' = 1/x`
`v' = 1/x, v = ln x`

`implies color{red}{int dx qquad ln(x)1/x} = (ln x)^2 - color{red}{int dx qquad ln x 1/x }+ C`

`implies 2 int dx qquad ln(x)1/x = (ln x)^2 + C`

`implies int dx qquad ln(x)1/x = 1/2 (ln x)^2 + C`