Solving $\int \frac{e^{x}+1}{e^{x}}$ with e Substitution

[dlthompson81]

Homework Statement

\begin{equation} \int \frac{e^{x}+1}{e^{x}}\end{equation}

Homework Equations

The Attempt at a Solution

This problem comes out of the substitution section of my book, and the answer in the back of the book is x-e^{x}+C

I started by changing the form to this:

\begin{equation}\int \frac{1}{e^{x}}e^{x}\end{equation}

I set u = e^{x} and du is the same.

That left me with:

\begin{equation}\int \frac{1}{u} du\end{equation}

Integrating \frac{1}{u} gives ln|u|

So plugging u back in gives ln|e^{x}|

I plugged some test numbers into my answer and the book answer and they come out pretty close, so I thought both of them may be right, but I still can't figure out how to get x-e^{-x}.

Any ideas?

 

[dynamicsolo]

Homework Statement

\begin{equation} \int \frac{e^{x}+1}{e^{x}}\end{equation}

For one thing, you changed the problem when you divided:

\frac{e^{x}+1}{e^{x}} = \frac{e^{x}}{e^{x}} + \frac{1}{e^{x}} = 1 + \frac{1}{e^{x}}

And I don't think you quite copied the answer from the book correctly...

 

[dlthompson81]

Ok, after reviewing it some more I found my problem. The derivative of e^{-x} is actually -e^{-x}. I didn't realize I had to bring down the negative sign.

And I did copy the answer wrong, it should be a -x exponent. My bad.

Thanks for the help.

 

[dlthompson81]

Ok, I was wrong again. I thought I had it figured out.

I have:

\begin{equation}\int 1 + e^{-x}\end{equation}

I set u = e^{-x} and du = -e^{-x} but then there is nothing for du to replace.

I'm lost again.

 

[Ivan92]

In this case, you would not do a u-sub. You would split up the integral (as I say it) like so:

\int1dx+\int e^{-x}dx

You should be able to integrate it from there.

 

[Mark44]

Ok, I was wrong again. I thought I had it figured out.

I have:

\begin{equation}\int 1 + e^{-x}\end{equation}

You need to get into the habit of including the differential, dx in this case. When you start out with integration techniques, as you are doing here, it's not so crucial. However, as the problems get more difficult, I guarantee[/color] that omitting the differential will come back and bite your hind end.

I set u = e^{-x} and du = -e^{-x} but then there is nothing for du to replace.

I'm lost again.

 

[dynamicsolo]

You need to get into the habit of including the differential, dx in this case. When you start out with integration techniques, as you are doing here, it's not so crucial. However, as the problems get more difficult, I guarantee[/color] that omitting the differential will come back and bite your hind end.

Besides the problems that can create for yourself, it is very likely that homework and exam graders will dock you on every problem where you omit it, or otherwise use notation incorrectly. (What you do in your own "scratch work" is your business, though being sloppy can cost you at some point...)

 

[dlthompson81]

In this case, you would not do a u-sub. You would split up the integral (as I say it) like so:

\int1dx+\int e^{-x}dx

You should be able to integrate it from there.

Yes, I can integrate it from here, but since the problem is under the substitution review section in my book, I thought I would try to figure out how to do it with substitution.

I usually write the dx whenever I work my problems out on my official paper, but I do avoid it on my scratch work. I guess I should break that habit.

 

[Mark44]

Yes, I can integrate it from here, but since the problem is under the substitution review section in my book, I thought I would try to figure out how to do it with substitution.

When you evaluate \int e^{-x}~dx, you will need to use a simple substitution.

I usually write the dx whenever I work my problems out on my official paper, but I do avoid it on my scratch work. I guess I should break that habit.

Yes, you should. When you get to trig substitutions, omitting the differential will definitely cause errors.