grrrrr another EZ integral

**TYTY** · March 7th 2009, 01:52 PM

$<br /> \int{\frac{e^{-x}}{1 + e^{-(2x)}}}<br />$

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.

**Scopur** · March 7th 2009, 01:58 PM

Split it into two different integrals..

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

EDIT: LaTeX fixed.

**Chris L T521** · March 7th 2009, 02:00 PM

Originally Posted by TYTY

$<br /> \int{\frac{e^{-x}}{1 + e^{-(2x)}}}<br />$

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 $\int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx$

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

Can you try the problem now?

**Chris L T521** · March 7th 2009, 02:02 PM

Originally Posted by Scopur

Split it into two different integrals..

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

You can't do this. You can't split the denominator up like that!!

**Scopur** · March 7th 2009, 02:07 PM

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

**Mentia** · March 7th 2009, 02:08 PM

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

**TYTY** · March 7th 2009, 02:11 PM

Originally Posted by Chris L T521

Note that $\int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx$

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

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...

**Mentia** · March 7th 2009, 02:18 PM

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.

**TYTY** · March 7th 2009, 02:21 PM

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

**Mentia** · March 7th 2009, 03:12 PM

Let me explain that one a little better.

We know:

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

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

Well, we can say,

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

Then

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

So then we use the chain rule:

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

So if we had some other constant, a, instead of 2:

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

So now you can see that this is the general form of a derivative of $a^{x}$

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