Answer
$$\frac{1}{6}{\tanh ^{ - 1}}\frac{{{e^x}}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{36 - {e^{2x}}}}dx} \cr
& {\text{rewriting the denominator}} \cr
& \int {\frac{{{e^x}}}{{{{\left( 6 \right)}^2} - {{\left( {{e^x}} \right)}^2}}}dx} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, }}\left| {{e^x}} \right|{\text{ > 0}} \cr
& \int {\frac{{dx}}{{{a^2} - {u^2}}}} = \frac{1}{a}{\tanh ^{ - 1}}\frac{u}{a} + C \cr
& ,so \cr
& = \frac{1}{6}{\tanh ^{ - 1}}\frac{{{e^x}}}{6} + C \cr} $$
