Question

Integrate e^3x-4/e^2x

Answer 1

Let u=e^x

Answer 2

\[\Large \int\limits_{}^{}e^{x} dx=e^{x}+c\]

Answer 3

Also \[\Large \int\limits\limits_{}^{}e^{ax} ~dx=\frac{1}{a}e^{ax}+c\]

Answer 4

In this case separate the two:
\[\Large e^{3x}\]
And the \[\Large -\frac{4}{e^{2x}}\]=\[\Large -4e^{-2x}\]

Answer 5

I did that and then set u=e^x but I don't know where to go from there.

Answer 6

I don't advise doing that. Set u=3x in the first one

Answer 7

Okay I did that and now what do I do?

Answer 8

\[\Large \int\limits_{}^{}e^{3x}-\frac{4}{e^{2x}}~~ dx=\int\limits_{}^{}e^{3x}~~ dx-4\int\limits_{}^{}e^{-2x}~~dx\]

Answer 9

Solve for dx when you take the differential of both sides

Answer 10

I set u=-2x and got -1/2du=dx. I don't know where to put the -1/2 though.

Answer 11

Plug it in for the dx

Answer 12

I get e^x-4e^u

Answer 13

\[\Large -4\int\limits\limits_{}^{}e^{-2x}~dx=~-4\int\limits\limits_{}^{}e^{u} \times \frac{ 1 }{ 2 }du\]

Answer 14

Simplifying, what do you get? We're simplifying the second equation first

Answer 15

-2e^udu?

Answer 16

There you go

Answer 17

OK and now do the integral based on that rule \[\Large \int\limits\limits_{}^{}e^{x} dx=e^{x}+C\]

Answer 18

-2e^-2x+c would be the answer.

Answer 19

Correct, now we can plug that into \[\Large \int\limits\limits_{}^{}e^{3x}-\frac{4}{e^{2x}}~~ dx=\int\limits\limits_{}^{}e^{3x}~~ dx-4\int\limits\limits_{}^{}e^{-2x}~~dx\]

Answer 20

\[\Large \int\limits\limits\limits_{}^{}e^{3x}-\frac{4}{e^{2x}}~~ dx=\int\limits\limits\limits_{}^{}e^{3x}~~ dx+2e^{-2x}+C\]

Answer 21

So is the final answer -1/2e^3x+2e^-2x+c?

Answer 22

There was a little mistake.
u= -2x
-1/2 du= dx

Answer 23

For the second one, you have to set u= 3x and do the same operation

Answer 24

@camila1315 are you still there/