Answer
$$\ln \left( {\frac{{{e^t} + 1}}{{{e^t} + 2}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^t}dt}}{{{e^{2t}} + 3{e^t} + 2}}} \cr
& {\text{Let }}u = {e^t},\,\,\,\,du = {e^t}dt \cr
& \int {\frac{{{e^t}dt}}{{{e^{2t}} + 3{e^t} + 2}}} = \int {\frac{{du}}{{{u^2} + 3u + 2}}} \cr
& \cr
& {\text{Decomposing the integrand }}\frac{1}{{{u^2} + 3u + 2}}{\text{ into partial fractions}} \cr
& \frac{1}{{{u^2} + 3u + 2}} = \frac{1}{{\left( {u + 2} \right)\left( {u + 1} \right)}} = \frac{A}{{u + 2}} + \frac{B}{{u + 1}} \cr
& {\text{Multiply by }}\left( {u + 2} \right)\left( {u + 1} \right) \cr
& 1 = A\left( {u + 1} \right) + B\left( {u + 2} \right) \cr
& \,\,\,\,\,\,\,\,{\text{If }}u = - 2,{\text{ then }}A = - 1 \cr
& \,\,\,\,\,\,\,\,{\text{If }}u = - 1,{\text{ then }}B = 1 \cr
& \frac{1}{{\left( {u + 2} \right)\left( {u + 1} \right)}} = \frac{A}{{u + 2}} + \frac{B}{{u + 1}} = - \frac{1}{{u + 2}} + \frac{1}{{u + 1}} \cr
& {\text{then}}{\text{,}} \cr
& \int {\frac{{du}}{{{u^2} + 3u + 2}}} = \int {\left( { - \frac{1}{{u + 2}} + \frac{1}{{u + 1}}} \right)du} \cr
& {\text{integrating}} \cr
& = - \ln \left| {u + 2} \right| + \ln \left| {u + 1} \right| + C \cr
& \cr
& {\text{write in terms of }}t;{\text{ replace }}{e^t}{\text{ for }}u \cr
& = - \ln \left| {{e^t} + 2} \right| + \ln \left| {{e^t} + 1} \right| + C \cr
& = \ln \left( {\frac{{{e^t} + 1}}{{{e^t} + 2}}} \right) + C \cr} $$