# Thread: grrrrr another EZ integral

1. ## grrrrr another EZ integral

$\displaystyle \int{\frac{e^{-x}}{1 + e^{-(2x)}}}$

I tried my very best to solve this but I am stuck. I really dislike my text book at all and my professor is very hard for me to understand. Does anyone have a good resource online for this type of problem? I understand all the new concepts we are supposed to be learning but I am stuck on the stuff I was supposed to learn last semester e.g. 5 years ago for me.

2. Split it into two different integrals..

$\displaystyle \int {e^{-x}} + \int{e^{-x}/e^{-2x}}$

EDIT: LaTeX fixed.

3. Originally Posted by TYTY
$\displaystyle \int{\frac{e^{-x}}{1 + e^{-(2x)}}}$

I tried my very best to solve this but I am stuck. I really dislike my text book at all and my professor is very hard for me to understand. Does anyone have a good resource online for this type of problem? I understand all the new concepts we are supposed to be learning but I am stuck on the stuff I was supposed to learn last semester e.g. 5 years ago for me.
Note that $\displaystyle \int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx$

I leave it for you to show that by the substitution $\displaystyle u=e^{-x}$, you have $\displaystyle \int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx\xrightarrow{u=e^{-x}}{}-\int\frac{\,du}{1+u^2}$

Can you try the problem now?

4. Originally Posted by Scopur
Split it into two different integrals..

$\displaystyle \int {e^{-x}} + \int{e^{-x}/e^{-2x}}$
You can't do this. You can't split the denominator up like that!!

5. Oh my bad.. i am not thinking well today apparently.. :S

6. use the substitution: u = e^(-x)

7. Originally Posted by Chris L T521
Note that $\displaystyle \int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx$

I leave it for you to show that by the substitution $\displaystyle u=e^{-x}$, you have $\displaystyle \int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx\xrightarrow{u=e^{-x}}{}-\int\frac{\,du}{1+u^2}$

Can you try the problem now?
your answer made this problem trivial - even to me haha. Weird thing is when I attempted to verify the derivative of e^(-x), I used google's first search result Online Derivative Calculator

Which says that
Function e^(-x)

f'x = -log(e)/e^x

pretty weird...

8. but what is the natural log of e? 1. So it gave you the right answer. It just didnt assume the "e" you entered is actually e but just some variable.

9. Originally Posted by Mentia
but what is the natural log of e? 1. So it gave you the right answer. It just didnt assume the "e" you entered is actually e but just some variable.
hm that makes sense. Given my level of frustration, I am getting nowhere fast with all this

10. Let me explain that one a little better.

We know:

$\displaystyle \frac{d }{dx } e^{x} = e^{x}$

But what about, say, $\displaystyle \frac{d }{dx } 2^{x}$?

Well, we can say,

$\displaystyle 2^{x} = (e^{ln(2)})^{x} = e^{xln(2)}$

Then

$\displaystyle \frac{d }{dx } 2^{x} = \frac{d }{dx }e^{xln(2)}$

So then we use the chain rule:

$\displaystyle \frac{d }{dx }e^{xln(2)} = ln(2)e^{xln(2)} = ln(2)2^{x}$

$\displaystyle \frac{d }{dx }e^{xln(a)} = ln(a)e^{xln(a)} = ln(a)a^{x}$
So now you can see that this is the general form of a derivative of $\displaystyle a^{x}$