Answer
$\frac{1}{4}\left( { - 3x - 2{e^{ - x}} + 3\ln \left| {2 + 3{e^x}} \right|} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{e^x}\left( {3{e^x} + 2} \right)}}} dx \cr
& {\text{Integrate by using a CAS }}\left( {{\text{See image below}}} \right){\text{ we obtain:}} \cr
& \int {\frac{1}{{{e^x}\left( {3{e^x} + 2} \right)}}} dx = \frac{1}{4}\left( { - 3x - 2{e^{ - x}} + 3\ln \left( {3{e^x} + 2} \right)} \right) + C \cr
& \cr
& {\text{Multiply the numerator and denominator by }}{e^x} \cr
& = \int {\frac{{{e^x}}}{{{e^{2x}}\left( {3{e^x} + 2} \right)}}} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& = \int {\frac{{du}}{{{u^2}\left( {3u + 2} \right)}}} \cr
& {\text{Integrating by tables using the entry 50}} \cr
& \int {\frac{{du}}{{{u^2}\left( {a + bu} \right)}}} = - \frac{1}{{au}} + \frac{b}{{{a^2}}}\ln \left| {\frac{{a + bu}}{u}} \right| + C \cr
& {\text{Let }}a = 2{\text{ and }}b = 3 \cr
& \int {\frac{{du}}{{{u^2}\left( {3u + 2} \right)}}} = - \frac{1}{{2u}} + \frac{3}{4}\ln \left| {\frac{{2 + 3u}}{u}} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}{e^x}{\text{ for }}u \cr
& \int {\frac{1}{{{e^x}\left( {3{e^x} + 2} \right)}}} dx = - \frac{1}{{2{e^x}}} + \frac{3}{4}\ln \left| {\frac{{2 + 3{e^x}}}{{{e^x}}}} \right| + C \cr
& {\text{Using the property of logarithms}} \cr
& = - \frac{1}{2}{e^{ - x}} + \frac{3}{4}\ln \left| {2 + 3{e^x}} \right| - \frac{3}{4}\ln \left| {{e^x}} \right| + C \cr
& = - \frac{1}{2}{e^{ - x}} + \frac{3}{4}\ln \left| {2 + 3{e^x}} \right| - \frac{3}{4}x + C \cr
& {\text{Factoring out }}\frac{1}{4} \cr
& = \frac{1}{4}\left( { - 2{e^{ - x}} + 3\ln \left| {2 + 3{e^x}} \right| - 3x} \right) + C \cr
& = \frac{1}{4}\left( { - 3x - 2{e^{ - x}} + 3\ln \left| {2 + 3{e^x}} \right|} \right) + C \cr
& {\text{We obtain the same result.}} \cr} $$
