find indefinite integral of: (6 - 3e^x) / (e^(2x))?
with working please, not sure how to do this integral, if possible solve with substitution because I understand that.
- 8 years agoBest Answer
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
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.
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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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- YirmiyahuLv 78 years ago
No need for substitution. It's just algebra:
(6 - 3e^x) / e^(2x) = 6/e^(2x) - 3e^x/e^(2x) = 6e^(-2x) - 3e^(-x)
Integrating these terms should be trivial.
- seguinLv 43 years ago
enable u=6-x^3 then you quite get u'=-3x^2 you have already got x^2 interior the equation so which you purely desire neg. 3, so multiply neg.3 interior the front of the crucial sign and multiply the finished element with the help of -a million/3 so as that they make one (so as that they cancel out) then you quite've crucial of u^a million/2= 2/3u^(3/2)=2/3(6-x^3)^(3/2) and that's the respond.