Answer
$20\ln|1-e^{-0.05x}|+C$
Work Step by Step
$I=\displaystyle \int\frac{e^{-0.05x}}{1-e^{-0.05x}}dx$
Substitute: $\left[\begin{array}{llll}
u & =1-e^{-0.05x} & & \\\\
du & =-(-0.05e^{-0.05x})dx & & \\
& =0.05e^{-0.05x}dx & \Rightarrow & e^{-0.05x}dx=\dfrac{du}{0.05}
\end{array}\right]$
$I=\displaystyle \int\frac{1}{u}\cdot\frac{du}{0.05}$
$=\displaystyle \frac{1}{0.05}\int u^{-1}du$
special case of the power rule $(n=-1)$
$=20\ln|u|+C$
bring back the variable $x$
$=20\ln|1-e^{-0.05x}|+C$
