# Find the integral

1. Jun 28, 2008

### jimen113

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution
u=e^x+1
du=e^x
=$$\int$$ u du
=(u$$^{}2$$/2)

2. Jun 28, 2008

### 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.

3. Jun 28, 2008

### jimen113

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)?

4. Jun 28, 2008

### jimen113

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

5. Jun 28, 2008

### rocomath

Why did it become a negative x?

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

6. Jun 28, 2008

### 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

7. Jun 28, 2008

### rocomath

You're breaking up the fraction incorrectly.

8. Jun 28, 2008

### jimen113

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

9. Jun 28, 2008

### 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}$$

Last edited: Jun 28, 2008
10. Jun 29, 2008

### dynamicsolo

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 $$e^{-x}$$?

Last edited: Jun 29, 2008
11. Jun 29, 2008

### 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?

12. Jun 29, 2008

### rocomath

And, yes that is the correct simplification.

13. Jun 29, 2008

### jimen113

x+(-e^-x) + C

14. Jun 29, 2008

### 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) ,