Question

integral of e^(sqrt(3s+9))ds.....? need help

Answer 1

Here's how I would do it, not sure if it's 100% right:

With exponent rules you have
\[e ^{\sqrt{3s+9}} = e ^{(3s+9)^{1/2}} = e ^{(3s+9)/2}\]

Use u-substitution:
Let u = \[(3s+9)/2\] and du = \[(3/2)ds\]

\[2/3\int\limits e ^{u}\]
= \[2/3e ^{u}+C\]
= \[2/3e ^{(3s+9)/2}+C\]
= \[2/3e ^{\sqrt{3s+9}}+C\]

Answer 2

I would do differently...Let U=3S+9 dU=3dS...

Resulting...\[1/3\int\limits_{}e^\sqrt{u}du\]

from there you require a P-substitution of the sqrt(u) term...Then you need to integrate by parts...
Ugly, messy, and no fun to do...Let WolframAlpha do some of the work....

http://www.wolframalpha.com/input/?i=integral+of+e^%28sqrt%283s%2B9%29%29ds

Enjoy.