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