## anonymous 5 years ago integral of e^(sqrt(3s+9))ds.....? need help

1. anonymous

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

2. anonymous

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.