Approach to this integral

**Jim Newt** · May 28th 2008, 07:58 AM

This needs to be integrated:

1/(e^(2x) + e^(x)) w/respect to x

Can't do partial fractions because e^x never equals zero. Right? Any suggesions? Thanks,

JN

**galactus** · May 28th 2008, 08:08 AM

I would go ahead and write it this way:

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

Now, it shouldn't be too bad.

$\int\frac{1}{e^{x}}dx=\int{e^{-x}}dx=-e^{x}$

The other can be done with a simple u sub:

**TKHunny** · May 28th 2008, 08:12 AM

You may wish to rethink that Partial Fraction idea. I knew there was something wrong with that method of finding Partial Fractions.

Have you considered $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} - \frac{1}{e^{x}+1}$ ?

**Jim Newt** · May 28th 2008, 09:00 AM

Originally Posted by TKHunny

You may wish to rethink that Partial Fraction idea. I knew there was something wrong with that method of finding Partial Fractions.

Have you considered $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} - \frac{1}{e^{x}+1}$ ?

Hey thanks for the input. But it seems that I'm having some sort of mathmatical disconnect here:

1. By factoring: $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} *\frac{1}{e^{x}+1}$
2. How does this equal $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} -\frac{1}{e^{x}+1}$ ?

**Moo** · May 28th 2008, 09:36 AM

Hello,

Originally Posted by Jim Newt

Hey thanks for the input. But it seems that I'm having some sort of mathmatical disconnect here:

1. By factoring: $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} *\frac{1}{e^{x}+1}$
2. How does this equal $\frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} -\frac{1}{e^{x}+1}$ ?

$\frac{1}{e^{x}(e^{x}+1)}=\frac{1+e^x-e^x}{e^{x}(e^{x}+1)}=\frac{1+e^x}{e^{x}(e^{x}+1)}-\frac{e^x}{e^{x}(e^{x}+1)}=\dots$

**Jim Newt** · May 28th 2008, 09:56 AM

Ah dang algebra! Thanks!

**Mathstud28** · May 29th 2008, 07:48 PM

Originally Posted by Jim Newt

This needs to be integrated:

1/(e^(2x) + e^(x)) w/respect to x

Can't do partial fractions because e^x never equals zero. Right? Any suggesions? Thanks,

JN

Just remember that if you get stuck you may legally do the following

Let $u=e^x$

So that

$\frac{1}{e^{2x}+e^x}=\frac{1}{e^x(e^x+1)}=\frac{1} {u(u+1)}$

Now let

$\frac{1}{u(u+1)}=\frac{A}{u}+\frac{B}{u+1}$

Multiplying through by $u(u+1)$ we get

$1=A(u+1)+B(u)$

Now letting $u=0$

we get

$1=A(0+1)+B(0)\Rightarrow{1=A}$

and letting $u=-1$

we see that

$1=A(1+-1)+B(-1)\Rightarrow{B=-1}$

so we see that

$\frac{1}{u(u+1)}=\frac{1}{u}-\frac{1}{u+1}$

now back subbing we see that

$\frac{1}{e^{2x}+e^x}=\frac{1}{e^x}-\frac{1}{e^x+1}$

and just to finish it

$\int\frac{dx}{e^x}+\int\frac{1}{e^x+1}dx=\int{e^{-x}dx}+\int\frac{e^x+1-e^x}{e^x+1}=$ $-e^{-x}+\int\frac{e^x+1}{e^x+1}dx-\int\frac{e^x}{e^x+1}dx=-e^{-x}+x-\ln|e^x+1|+C$

Thread: Approach to this integral

LinkBack

Thread Tools

Display

Approach to this integral

Similar Math Help Forum Discussions

Which approach is nicer? UK MEI approach? American approach?

Any approach other than axiomatic approach to Real Numbers

How should i approach this integral question?

How can I approach this integral???

I'm not sure how to approach this...

Search Tags