how do you turn this?

Question

I= - [ 1/x lnx + 1/x ] from e to 7 (this is and integration prob)
turn that into this answer:
I= [(2/e) - (1/7)] (ln7+1)

Answer 1

7
∫ - [ 1/(x lnx) + 1/x ]dx =
e
- [ln(ln(x)) + ln(x)] =
- ln(ln(7)) - ln(7) + ln(ln(e) + ln(e) =
1 - ln(ln(7)) - ln(7)

7
∫ - [ (1/x) lnx + 1/x ]dx =
e
- ln(7)((1/2)ln(7) + 1) + ln(e)((1/2)ln(e) + 1) =
- ln(7)((1/2)ln(7) + 1) + 3/2

If the integration is already complete,
I = - [(1/7)ln(7) + 1/7] + (1/e)ln(e) + 1/e
I = (2/e) - (1/7)(ln(7) + 1)

I believe that this is as close to your given answer as you can get.