12.1. Calculate: $\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{\sqrt{e^{x}+e^{-x}}} d x$.
Solution:

$$
\begin{gathered}
\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{\sqrt{e^{x}+e^{-x}}} d x=\int_{0}^{\ln 2} \frac{\left(e^{x}-e^{-x}\right)\left[\left(e^{x}+e^{-x}\right)^{2}-1\right]}{\sqrt{e^{x}+e^{-x}}} d x=\left|\begin{array}{c}
\sqrt{e^{x}+e^{-x}}=t \\
e^{x}+e^{-x}=t^{2} \\
\left(e^{x}-e^{-x}\right) d x=2 t d t \\
x=0 \Rightarrow t=\sqrt{2} \\
x=\ln 2 \Rightarrow t=\frac{1}{2} \sqrt{10}
\end{array}\right|= \\
=2 \int_{\sqrt{2}}^{\frac{1}{2} \sqrt{10}}\left(t^{4}-1\right) d t=\left.2\left(\frac{t^{5}}{5}-t\right)\right|_{\sqrt{2}} ^{\frac{1}{2} \sqrt{10}}=\frac{5 \sqrt{10}+8 \sqrt{2}}{20}
\end{gathered}
$$