Example 2.57. Calculate the length of the arc of the curve $x=0.25 y^{2}-$ $-0.5 \ln y$, enclosed between the points $y=1$ and $y=1.5$.
Solution. According to formula (2.25)

$$
\begin{aligned}
& L=\int_{1}^{1.5} \sqrt{1+\left(0.5 y-\frac{0.5}{y}\right)^{2}} d y=\int_{1}^{1.5} \sqrt{\left(0.5 y+\frac{0.5}{y}\right)^{2}} d y= \\
& =\int_{1}^{1.5}\left(0.5 y+\frac{0.5}{y}\right)^{2} d y=\left.(y-0.5 \ln |y|)\right|_{1} ^{1.5}=0.5(1-\ln 1.5) .
\end{aligned}
$$