Answer
$\displaystyle \frac{1}{3}\ln|\frac{2e^{x}-1}{e^{x}+1}|+C$
Work Step by Step
I=$\displaystyle \int\frac{dx}{1+2e^{x}-e^{-x}}=\quad \left[\begin{array}{lll}
u=e^{x} & & x=\ln u\\
du=e^{x}dx & & dx=\frac{du}{u}
\end{array}\right]$
$=\displaystyle \int\frac{\frac{du}{u}\times u}{(2u+1-\frac{1}{u})\times u}=\int\frac{du}{2u^{2}+u-1}$
Factor the denominator
$2u^{2}+u-1=2u^{2}+2u-u-1$
$= 2u(u+1)-(u+1)=(2u-1)(u+1)$
$I=\displaystyle \int\frac{du}{(2u-1)(u+1)}$
(use the partial fractions method)
$\displaystyle \frac{1}{(2u-1)(u+1)}=\frac{A}{2u-1}+\frac{B}{u+1}$
$1=Au+A+2Bu-B$
$1=(A+2B)u+(A-B)\Rightarrow\left\{\begin{array}{l}
A+2B=0\\
A-B=1
\end{array}\right.$
Subtract the second from the first: $3B=-1,$
$B=-\displaystyle \frac{1}{3} .\quad A=\frac{2}{3}$
$I=\displaystyle \frac{2}{3}\int\frac{du}{2u-1}-\frac{1}{3}\int\frac{du}{u+1}$
(table of integrals, type 2)
$=\displaystyle \frac{2}{3}\cdot\frac{1}{2}\ln|2u-1|-\frac{1}{3}\ln|u+1|+C$
$=\displaystyle \frac{1}{3}(\ln|2u-1|-\ln|u+1|)+C$
(bring back x, apply logarithm of a quotient)
$=\displaystyle \frac{1}{3}\ln|\frac{2e^{x}-1}{e^{x}+1}|+C$
