Question

Evaluate the integral using integration by parts where possible.
((2x+1)/(e^9x))dx

Answer 1

\[\int\limits_{}^{}\frac{2x+1}{e^{9x}} dx=\int\limits_{}^{}(\frac{2x}{e^{9x}}+\frac{1}{e^{9x}}) dx\]
\[=\int\limits_{}^{}2xe^{-9x} dx+\int\limits_{}^{}e^{-9x} dx\]

Answer 2

Does that form look better to you? Or are you still uncertain on how to evaluate this?

Answer 3

It's a much better form, so is the final answer the actual final answer or does it need more work to it?

Answer 4

You to actually integrate
You know use integration by parts to integrate the first integral
And then the second integral is tons easier

Answer 5

Just so I know I'm on the right track...e^-9x integrated is still e^-9x right? Or does that require u substitution.

Answer 6

use this for the second part!
\[\int\limits_{}^{}e^{c \cdot x} dx=\frac{1}{c} e^{c \cdot x} +C \text{ where c is a constant not equal to zero}\]

Answer 7

For the first part...
You can do a table

Answer 8

So 1/-9 e^-9x + C? For the second part. The first part I know how to do.

Answer 9

Yes that is all for the second part

Answer 10

Okay cool, thank you very much!

Answer 11

So you know how to do the first part which is what I find weird since you didn't know how to do the second part

Answer 12

Are you sure you know how to do the first part?

Answer 13

You want to tell me what you got for the answer? I will check.

Answer 14

I can tell you what I get after I finish it if you'd like.

Answer 15

That is fine :)

Answer 16

Because you told me how to do the second part I can do the first part, which is why I didn't need help with the first part.

Answer 17

lol Oh okay. Gotcha!

Answer 18

I'm not really sure if this is right, but i used the formula I u dv = u*v - I v*dv, where I stands for integral.
Here's what I got:
(2x)((-1/9)e^-9x)+(1/9)-((1/9e^-9x))*((1/9e^-9x))+C
It's not simplified at all I realize.

Answer 19

Ok it is hard for me to read that but if you are saying what I got here, then you are correct:
\[\int\limits_{}^{}fg'dx=fg-\int\limits_{}^{}f'g dx\]
\[f=2x, g'=e^{-9x} => f'=2, g=\frac{-1}{9} e^{-9x}\]
So plugging into our formula we get:
\[\int\limits_{}^{}2x \cdot e^{-9x} dx=2x \cdot \frac{-1}{9}e^{-9x}-\int\limits_{}^{} 2 \cdot \frac{-1}{9}e^{-9x} dx\]
\[=\frac{-2}{9}xe^{-9x}+\frac{2}{9} \cdot \frac{-1}{9}e^{-9x} +C \]
\[=\frac{-2}{9}xe^{-9x}-\frac{2}{81}e^{-9x}+C\]
Is that what you have for the first part?

Answer 20

I think you are off by a constant multiply in your second term

Answer 21

multiple*

Answer 22

Looks like you are missing the constant multiple 2

Answer 23

just for the second term if i read right

Answer 24

What you have is a simplified version of what I had, which is what I didn't know how to do.

Answer 25

wait you have e^{-9x}*e^{-9x}
you should only have this factor once not twice for the second term

Answer 26

I do, I just combined the first and second parts so I could get an entire answer.

Answer 27

I'm not sure what you are saying, but for the first part we have
\[\frac{-2}{9}xe^{-9x}-\frac{2}{81}e^{-9x}+C_2\]
And the second part we have
\[\frac{-1}{9}e^{-9x}+C_1\]

Do you understand how I got or we got both of these parts?

Answer 28

Yeah, but to answer the original question, part 1 and part 2 have to be combined. You helped me tremendously!

Answer 29

That is correct we need to add the parts

Answer 30

\[\frac{-2}{9}xe^{-9x}-\frac{2}{81}e^{-9x}+\frac{-1}{9}e^{-9x}+C\]

Answer 31

I got the answer right! :) Thanks so much for your help!

Answer 32

Ok great! Have a nice day.

Answer 33

Thanks, you too!