# Evaluating Integral with a natural log

1. Sep 1, 2008

### XedLos

1. The problem statement, all variables and given/known data
$$\int9s9^s ds$$

2. Relevant equations
∫udv=uv-∫9^sds-∫vdu

3. The attempt at a solution
u=9s
du=9ds

dv=9^s ds
v=∫9^sds
=∫3^(2s) ds
=3^(2s)/[2ln(3)]
this is as far as i have gotten. Am i correct so far?

Last edited: Sep 1, 2008
2. Sep 2, 2008

### HallsofIvy

Staff Emeritus
Yes, that is correct. You didn't really need to reduce to 3, that is exactly the same as
$$\int 9^s ds= 9^s/ln(9)$$