# Revenge of the Integrals

• Apr 26th 2006, 02:33 AM
Sophie
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!
• Apr 26th 2006, 03:34 AM
CaptainBlack
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!

$
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}
$
,

and from there the rest should be easy.

RonL
• Apr 26th 2006, 06:44 PM
Jameson
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.