Best Answer:
Hi John,

First thing to note is that we can break this guy into separate integrals:

(6 - 3e^x)/e^(2x) = 6/e^(2x) - 3e^(x-2x) = 6e^(-2x) - 3e^(-x), and now you can solve as two separate integrals:

(1) 6e^(-2x); and

(2) 3e^(-x).

To integrate (1), use a "u" substitution: u = -2x with du = -2dx. This transforms (1) into:

Integral(6e(-2x),x) = Integral(-3e^u,u) = -3e^u + c = -3e^(-2x) + c.

Likewsie in (2), let u = -x with du = -dx:

Integral(3e^(-x),x) = Integral(-3e^u,u) = -3e^u + c = -3e^(-x) + c.

Subtracting (1) - (2), means our answer is:

-3e^(-2x) + 3e^(-x) + C.

To check, differentiate:

Dx[-3e^(-2x) + 3e^(-x) + C] = 6e(-2x) - 3e(-x), which is what we started with. We can also check with the Wolfram online integrator, which is referenced below.

To help reinforce your understanding of the solving indefinite integrals with exponential functions, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.

As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

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Source(s):
http://www.youtube.com/watch?v=DffQa5SyMSE

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/expondirectory/Exponentials.html

http://integrals.wolfram.com/index.jsp

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