Answer
$\displaystyle \frac{e^{t}}{2}\sqrt{e^{2t}-1}+\frac{1}{2}\ln|e^{t}+\sqrt{e^{2t}-1}|+C$
Work Step by Step
With the substitution $\left[\begin{array}{ll}
u=e^{t} & du=e^{t}dt\\
e^{2t}=u^{2}, & e^{3t}=u^{2}du
\end{array}\right]$
the integral becomes
$\displaystyle \int\frac{u^{2}}{\sqrt{u^{2}-1^{2}}}$ , for which we use formula
44. $\displaystyle \int\frac{u^{2}du}{\sqrt{u^{2}-a^{2}}}=\frac{u}{2}\sqrt{u^{2}-a^{2}}+\frac{a^{2}}{2}\ln|u+\sqrt{u^{2}-a^{2}}|+C$
bringing back the variable t
$=\displaystyle \frac{e^{t}}{2}\sqrt{e^{2t}-1}+\frac{1}{2}\ln|e^{t}+\sqrt{e^{2t}-1}|+C$
