Answer
$$\frac{{{e^{2t}}}}{2} + \frac{1}{2}\ln \left( {{e^{2t}} + 1} \right) - {\tan ^{ - 1}}\left( {{e^t}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{4t}} + 2{e^{2t}} - {e^t}}}{{{e^{2t}} + 1}}} dt \cr
& {\text{Factor the numerator}} \cr
& = \int {\frac{{\left( {{e^{3t}} + 2{e^t} - 1} \right){e^t}}}{{{e^{2t}} + 1}}} dt \cr
& = \int {\frac{{{e^{3t}} + 2{e^t} - 1}}{{{e^{2t}} + 1}}} \left( {{e^t}dt} \right) \cr
& {\text{let }}x = {e^t},\,\,\,dx = {e^t}dt \cr
& {\text{Write the integral in terms of }}x \cr
& \int {\frac{{{e^{3t}} + 2{e^t} - 1}}{{{e^{2t}} + 1}}} \left( {{e^t}dt} \right) = \int {\frac{{{x^3} + 2x - 1}}{{{x^2} + 1}}} dx \cr
& {\text{Using long division}} \cr
& \int {\frac{{{x^3} + 2x - 1}}{{{x^2} + 1}}} dx = \int {\left( {x + \frac{{x - 1}}{{{x^2} + 1}}} \right)} dx \cr
& = \int {\left( {x + \frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}}} \right)} dx \cr
& {\text{Distribute}} \cr
& = \int x dx + \int {\frac{x}{{{x^2} + 1}} - \int {\frac{1}{{{x^2} + 1}}} dx} \cr
& {\text{integrating}} \cr
& = \frac{{{x^2}}}{2} + \frac{1}{2}\ln \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + C \cr
& \cr
& {\text{Write the solution in terms of }}x,{\text{ substitute }}{e^t}{\text{ for }}x \cr
& = \frac{{{{\left( {{e^t}} \right)}^2}}}{2} + \frac{1}{2}\ln \left( {{{\left( {{e^t}} \right)}^2} + 1} \right) - {\tan ^{ - 1}}\left( {{e^t}} \right) + C \cr
& = \frac{{{e^{2t}}}}{2} + \frac{1}{2}\ln \left( {{e^{2t}} + 1} \right) - {\tan ^{ - 1}}\left( {{e^t}} \right) + C \cr} $$