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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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.

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