Question

Integration by parts.

integral of 7x/(e^x) dx

Answer 1

Changing it to 7x * e^(-x) dx

Answer 2

u = 7x
1/7 du = dx

dv = e^(-x)

v = -e^(-x)

Answer 3

\[7x *-e^{-x} +\frac{1}{7} \int\limits e^{-x} dx\]

Answer 4

\[7x *-e^{-x} -\frac{1}{7} e^{-x} +c?\]

Answer 5

Apparently it's not right....hmm damnit....

Answer 6

let me try

Answer 7

It says that it may not need to be used by integration by parts, but I think this has to be.

Answer 8

We have
\[\int 7x\times e^{-x} dx\]
We do need to use by parts, can't be done without it.

Answer 9

yup.

Answer 10

Check what I did above.... I'm not sure whereI messed up...

Answer 11

\[\int 7x\times e^{-x} dx\]
We know by parts
\[\int u v=u\int v-\int (du \int v)\]
\[\int 7x \times e^{-x} dx= 7x\int e^{-x} dx-\int (\frac{d}{dx} 7x \int e^{-x} dx)dx\]
we get
\[-7xe^{-x}-(\int 7\times -e^{-x} dx)\]

\[-7xe^{-x} +\int 7e^{-x} dx\]
we get
\[-7xe^{-x}-7e^{-x} +c\]

Answer 12

Do you get this?

Answer 13

Oh I fluttered up by doing the 1/7 huh I think I had this konfused before as well.

Answer 14

because we aren't switching to the u "world" we are staying with x....

Answer 15

oops, do you get it now?

Answer 16

Yeah I keep konfusing myself with this I guess :P.

Answer 17

We keep it as du = 7dx :P.

Answer 18

not the other way around, thanks :D

Answer 19

try to keep things simple, if we had \( e^{7x}\), then we'd need substitution. But still it's not necessary:)

Answer 20

:)

Answer 21

well I was just konfused because normally we are working with dx and then going to udu or w/e, but now we take du and switch back. I guess it depends on the situaiton tho....

Answer 22

yeah, here substitution is not of any help.