Thread: Approach to this integral

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

I would go ahead and write it this way:

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

Now, it shouldn't be too bad.

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

The other can be done with a simple u sub:

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

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

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

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

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

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

Ah dang algebra! Thanks!

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 $\displaystyle u=e^x$

So that

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

Now let

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

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

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

Now letting $\displaystyle u=0$

we get

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

and letting $\displaystyle u=-1$

we see that

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


so we see that

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

now back subbing we see that

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


and just to finish it

$\displaystyle \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}=$$\displaystyle -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$