Question

Find the indefinite integral:

Integral of (e^(-x))(6+(e^(-x)) dx

Answer 1

Which part would be u?

Answer 2

is it the 6+e^(-x) part?

Answer 3

Yep!

Answer 4

You hardly need a u-sub here. Just multiply out, and integrate each part separately. \[\int\limits e^{-x}(6+e^{-x})\;\;dx\]\[=\int\limits 6e^{-x} \;\;dx +\int\limits e^{-x}\cdot e^{-x}\;\;dx\]\[=\int\limits 6e^{-x} \;\;dx +\int\limits e^{-2x}\;\;dx\]

Answer 5

ooohhh, see that's what I was kind of thinking, but the section is has got me so stuck on subbing u.. lol

Answer 6

Well, \(u=6+e^{-x}\) would also work.

Answer 7

so.. a general question. When is it good to substitute and when is it better to just distribute and solve?

Answer 8

I would use u=-x
du=-dx
\[-\int\limits e^u \left(e^u+6\right) \, du\]
then sub again h=e^u
dh=e^u du

\[-\int\limits (h+6) \, dh\]
\[-\int\limits 6 \, dh-\int\limits h \, dh\]
\[-\frac{h^2}{2}-6 h+c\]
substitube back
\[-\frac{1}{2} e^{-2 x} \left(12 e^x+1\right)+c\]

Answer 9

My opinion is set priority for separate them out first!
If you can't, think the next step substitution!

Answer 10

@KingGeorge - so i could take the integral from the multiplied out format

Answer 11

@Chlorophyll thanks, that's a good rule to go by to know which to use

Answer 12

You could definitely do that. It would be my preferred way, but only because I don't like doing u-subs very much.

Answer 13

thanks everyone! makes more sense now :)

Answer 14

You're welcome.

Answer 15

Pretty much like if you can't substitute, then next step is Integral by Part :)

Answer 16

And if you can't do by by parts... Something has gone horribly horribly wrong.

Answer 17

gotcha.. you all are such lifesavers!

Answer 18

Then trig subs :P

Answer 19

ewwwww no talk about trig! lol ;)

Answer 20

I would say that if you have to do trig subs, something has already gone horribly horribly wrong.

Answer 21

lol, noted @KingGeorge ! :)

Answer 22

Yup, that's why I mean! However sb is so fond of trig, to the point apply it instead of U sub simple exponent var :P

Answer 23

so the integral of e^-x would be -e^-x?

Answer 24

*what :) Glad that majority feel the same =)

Answer 25

Yep, careful with negative !

Answer 26

and then e^(-2x) would turn into -e^(-2x)?

Answer 27

natural log gets me mixed up sometimes, that's why i'm checking

Answer 28

You've got to bring the 2 down as well. So you should get -2e^(-2x)

Answer 29

oohhh ok right.. thank you

Answer 30

e^u = u' e^u
u = -2x --> u' = -2

Answer 31

and my answer would be:
12(e^(-x))+(e^(-2x))

Answer 32

oh, and +C at the end

Answer 33

I'm getting something slightly different.

Answer 34

If you divided by 2 it would be correct. Go over your work and make sure you've accounted for everything like that.

Answer 35

ok

Answer 36

Divided by -2 actually.

Answer 37

I got

-6(integrand) of e^(-x) + -2(integrand) of e^(-2x), then I combined the contstants.. maybe those aren't combinable?