## Problem Statement

Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.

$$
y=\ln \frac{5}{2 x}, \sqrt{3} \leq x \leq \sqrt{8}
$$
## Solution

The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula

$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$

Let's find the derivative of the given function:

$$
f^{\prime}(x)=\left(\ln \frac{5}{2 x}\right)^{\prime}=\left(\ln \frac{5}{2}-\ln x\right)^{\prime}=-\frac{1}{x}
$$

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{\sqrt{3}}^{\sqrt{8}} \sqrt{1+\frac{1}{x^{2}}} d x= \\
& =\int_{\sqrt{3}}^{\sqrt{8}} \frac{\sqrt{1+x^{2}}}{x} d x=\left|\begin{array}{r}
\sqrt{1+x^{2}}=t ; x=\sqrt{t^{2}-1} \\
d x=\frac{t d t}{\sqrt{t^{2}-1}} \\
t_{1}=\sqrt{3+1}=2 ; t_{2}=\sqrt{8+1}=3
\end{array}\right|=\int_{2}^{3} \frac{t}{\sqrt{t^{2}-1}} \cdot \frac{t d t}{\sqrt{t^{2}-1}}= \\
& =\int_{2}^{3} \frac{t^{2}-1+1}{t^{2}-1} d t= \\
& =\int_{2}^{3} d t+\int_{2}^{3} \frac{d t}{t^{2}-1}= \\
& =\left.t\right|_{2} ^{3}+\left.\frac{1}{2} \ln \left|\frac{t-1}{t+1}\right|\right|_{2} ^{3}=3-2+\frac{1}{2} \ln \frac{1}{2}-\frac{1}{2} \ln \frac{1}{3}=1+\frac{1}{2} \ln \frac{3}{2}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-4$ »

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

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## Kuznetsov's Problem Book Integrals 17-5

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