2.2. Find the length of the arc $A B$ of the curve $e^{2 y}\left(e^{2 x}-1\right)=e^{2 x}+1$, if $x_{A}=1, x_{B}=2$.
Solution. The length of a smooth curve is calculated using a line integral of the first kind with the integrand function $f(x, y)=1$. First, let's find the explicit equation of the given curve:

$$
y=\frac{1}{2} \ln \frac{e^{2 x}+1}{e^{2 x}-1}
$$

Using this formula, we reduce the line integral to a definite integral:

$$
\begin{gathered}
\int_{A B} d l=\int_{1}^{2} \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\int_{1}^{2} \sqrt{1+\frac{4 e^{4 x}}{\left(e^{4 x}-1\right)^{2}}} d x=\int_{1}^{2} \frac{e^{2 x}+1}{e^{2 x}-1} d x= \\
\quad=\int_{1}^{2}\left(1+\frac{2}{e^{4 x}-1}\right) d x=1+2 \int_{1}^{2} \frac{d x}{e^{4 x}-1}=1+\frac{1}{2} \int_{e^{4}-1}^{e^{8}-1} \frac{d t}{t(t+1)}= \\
=1+\frac{1}{2} \int_{e^{4}-1}^{e^{8}-1} \frac{d t}{t}-\frac{1}{2} \int_{e^{4}-1}^{e^{8}-1} \frac{d t}{t+1}=1+\left.\frac{1}{2} \ln \frac{t}{t+1}\right|_{e^{4}-1} ^{e^{8}-1}=\frac{1}{2} \ln \left(1+e^{4}\right)-1
\end{gathered}
$$