You can do this just like you would a partial fractions problem. Think of $\frac{1}{e^{2t}(e^t+1)}$ as a rational function of the form $\frac{1}{u^2(u+1)}.$ Doing partial fractions, we have: $\begin{align*} \frac{1}{e^{2t}(e^t+1)} &= \frac{A}{e^t} + \frac{B}{e^{2t}} + \frac{C}{e^{t}+1}\\ &= \frac{Ae^t(e^t+1) + B(e^t+1) + Ce^{2t}}{e^{2t}(e^t+1)}\\ &= \frac{(A+C)e^{2t} + (A+B)e^t + B}{e^{2t}(e^t+1)}. \end{align*}$ So we want $A+C=0$, $A+B=0$, and $B=1$. Therefore, $A=-1$, $C=1$. Therefore, $\begin{align*} \int\frac{1}{e^{2t}(e^t+1)}\,dt &= \int\left(\frac{-1}{e^t} + \frac{1}{e^{2t}} + \frac{1}{e^t+1}\right)\,dt\\ &= -\int e^{-t}\,dt + \int e^{-2t}\,dt + \int\frac{1}{e^t+1}\,dt. \end{align*}$ The first two integrals are an easy substitution. The third integral may take a bit more doing, but setting $u=e^t+1$, $du=e^t\,dt$ gives $dt = \frac{1}{u-1}\,du$, so you can do another partial fraction to get $\int\frac{dt}{e^t+1} = \int\frac{du}{(u-1)u} = \int\left(\frac{1}{u-1} - \frac{1}{u}\right)\,du$ which is easy to do, yielding the final answer once everything is done.