Question

How do you calculate `int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx` ?

Answer 1

`int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx=ln(x)^ln(x)+C`

Explanation:

`\ \ \ \ \ \ int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx`

Let's first substitute `u=ln(x)` and `du=dx/x`:
`=int\ u^u(1+ln(u))\ du`
`=int\ e^(u ln(u))(1+ln(u))\ du`

Substitute again with `v=u ln(u)` and `dv=(1+ln(u))\ du`:
`=int\ e^v\ dv`
`=e^v+C`

Substitute back `v=u ln(u)`:
`=e^(u ln(u))+C`
`=u^u+C`

Substitute back `u=ln(x)`:
`=ln(x)^ln(x)+C`