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Math Help - Revenge of the Integrals

  1. #1
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    Revenge of the Integrals

    Just when I thought I was getting good at integration and logarithms, I encountered this question:

    Find indefinite integral of 3^(2x-1)

    I'm sure the method is quite simple-- I just want to know how to get to the solution [3^(2x-1)]/2ln3. Any help would be greatly appreciated!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Sophie
    Just when I thought I was getting good at integration and logarithms, I encountered this question:

    Find indefinite integral of 3^(2x-1)

    I'm sure the method is quite simple-- I just want to know how to get to the solution [3^(2x-1)]/2ln3. Any help would be greatly appreciated!
    <br />
3^{2x-1}=[e^{\ln(3)}]^{2x-1}=e^{\ln(3) (2x-1)}=e^{\ln(3)}e^{2\ln(3) x}=3\ e^{2\ln(3) x}<br />
,

    and from there the rest should be easy.

    RonL
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  3. #3
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    This isn't as rigorous, but think of integration in terms of the opposite of differentiation.

    Take the expression y=3^x. Remember that \frac{dy}{dx}=3^x*\ln{(3)} We multiply by the derivative when differentiating, so we must divide by it when integrating. \int 3^xdx=\frac{3^x}{\ln{(3)}}+C. Again, not rigorous but it might help visualize the method.
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