What is the Integral of (e^x + 1)/e^x?

[jimen113]

Homework Statement

\int (e^{}x + 1)/e^{}x dx

Homework Equations

the answer is: x-e^{}-x + C
after I integrate I don't get the above answer.

The Attempt at a Solution

u=e^x+1
du=e^x
=\int u du
=(u^{}2/2)

 

[nicksauce]

Well your method is wrong. If you let u = e^x + 1, du = e^xdx, then you have udu = (e^x+1)(e^xdx), which is clearly different from (e^x+1)/(e^x).

Hint: (a+b)/a = 1 + b/a.

 

[jimen113]

Thanks for your help.
So,

\int u/du
I don't know how to start integrating this fraction (that's if this is correct).
Integrate the top (u) and then the bottom (du)?

 

[jimen113]

Is this correct?
\int U/du=
e^{}x /e^{}x + 1
=1/e^{}x + X
=e^{}-x - X

 

[rocomath]

Why did it become a negative x?

\int(e^{-x}+1)dx

 

[jimen113]

I'm sorry, I should have typed:
=1/(e^{}x + x)
and when I brought it up to get rid of the fraction, then it became (e^{}-x -x

 

[rocomath]

I'm sorry, I should have typed:
=1/(e^{}x + x)
and when I brought it up to get rid of the fraction, then it became (e^{}-x -x

You're breaking up the fraction incorrectly.

 

[jimen113]

? confused, I will review the problem again. Any suggestions?
Thanks.

 

[rocomath]

\int(e^{-x}+1)dx=\pm e^{-x}\pm x+C

What should be the proper signs? positive or negative for e^(-x) and x ?

What's incorrect with the arithmetic is that ... \frac{1}{e^x+x}\neq\(e^{-x}-x ... it would be \frac{1}{e^x+x}=(e^{x}+x)^{-1}

 

[dynamicsolo]

Is this correct?
\int U/du= ...

One other problem with this is that du can never be in the denominator in an integration. Keep in mind that integration is the inverse of differentiation and that

F(x) = \int F'(x) dx = \int \frac{dF}{dx} dx = \int dF

dF being the "differential" of F. Having the differential dx in the denominator does not make sense in terms of the meaning of the symbols for integration.

As for the rest of post #4, I think you were on the right track, but then confused matters. It looks like you were finding the quotient and integrating it at the same time. Maybe we should go back and ask first: what does \frac{e^{x} + 1}{e^x} simplify to?

As another question relevant to the solution, what is the anti-derivative of e^{-x}?

 

[jimen113]

Looks like I'm an idiot, but I'm trying to understand this "integration" concept. I appreciate your help.

what does \frac{e^{x} + 1}{e^x} simplify to?
simplified to: 1+(1/e^x)
integrating above= x-(e^-x)
seems like the answer in the solutions manual:
x-e^-x+c

Is this process correct?

 

[rocomath]

Looks like I'm an idiot, but I'm trying to understand this "integration" concept. I appreciate your help.

what does \frac{e^{x} + 1}{e^x} simplify to?
simplified to: 1+(1/e^x)
integrating above= x-(e^-x)
seems like the answer in the solutions manual:
x-e^-x+c

Is this process correct?

Ok, so who cares about the answer in the SM. What is YOUR answer? :)

And, yes that is the correct simplification.

 

[jimen113]

Ok, so who cares about the answer in the SM. What is YOUR answer? :)

my answer is:
x+(-e^-x) + C

 

[dynamicsolo]

There you go! And you can always check the result of an integration (not always easy, but particularly important when your SM doesn't include an answer) by differentiating it:

d/dx [ x - (e^-x) + C]

= 1 - [ (e^-x)·{ d/dx (-x) } ] + 0

= 1 - (-1)·(e^-x) = 1 + e^-x ,

or, multiplying by (e^x)/(e^x) ,

[ { 1 + (e^-x) } · e^x ] / (e^x) = ( {e^x} + 1 ) / (e^x) ,

your original integrand.

And, no, you're not "an idiot". Just review some of your algebra and the definition of integration... and keep practicing!