Question

Integrals,

How would you solve this integral?

http://screencast.com/t/KozJYtHwSN7w

Answer 1

Do you remember this derivative?\[\Large (3^u)'\quad=\quad ?\]

Answer 2

no :(

Answer 3

D:

Answer 4

\[\Large (e^u)'\quad=\quad e^u\]When our base is anything besides e, we get an extra factor of `the natural log of the base`.\[\Large (3^u)'\quad=\quad 3^u(\ln3)\]
You can do some logarithmic differentiation if you want to justify that. ^

Answer 5

With integration, we end up `dividing` by that extra factor instead of multiplying.

Answer 6

\[\Large \int\limits 2^u\;du \quad=\quad \frac{2^u}{\ln2}\]

Answer 7

ty