Answer
$$ = \ln \left( {\frac{{{e^t} + 1}}{{{e^t} + 2}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^t}dt}}{{{e^{2t}} + 3{e^t} + 2}}} \cr
& {\text{let }}x = {e^t},\,\,\,dx = {e^t}dt \cr
& {\text{Write the integral in terms of }}x \cr
& \int {\frac{{{e^t}dt}}{{{e^{2t}} + 3{e^t} + 2}}} = \int {\frac{{dx}}{{{x^2} + 3x + 2}}} \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{x^2} + 3x + 2}} = \frac{1}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = \frac{A}{{x + 2}} + \frac{B}{{x + 1}} \cr
& {\text{Multiplying by }}\left( {x + 2} \right)\left( {x + 1} \right){\text{, we have}} \cr
& {\text{1}} = A\left( {x + 1} \right) + B\left( {x + 2} \right) \cr
& {\text{if we set }}x = - 2, \cr
& {\text{1}} = A\left( { - 1} \right) \cr
& A = - 1 \cr
& {\text{if we set }}x = - 1, \cr
& {\text{1}} = B\left( 1 \right) \cr
& B = 1 \cr
& {\text{then}} \cr
& \frac{1}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = \frac{A}{{x + 2}} + \frac{B}{{x + 1}} = \frac{{ - 1}}{{x + 2}} + \frac{1}{{x + 1}} \cr
& \int {\frac{{dx}}{{{x^2} + 3x + 2}}} = \int {\left( { - \frac{1}{{x + 2}} + \frac{1}{{x + 1}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - \ln \left| {x + 2} \right| + \ln \left| {x + 1} \right| + C \cr
& = \ln \left| {\frac{{x + 1}}{{x + 2}}} \right| + C \cr
& {\text{Write the solution in terms of }}x,{\text{ substitute }}{e^t}{\text{for }}x \cr
& = \ln \left| {\frac{{{e^t} + 1}}{{{e^t} + 2}}} \right| + C \cr
& = \ln \left( {\frac{{{e^t} + 1}}{{{e^t} + 2}}} \right) + C \cr} $$
