## Problem Statement

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

$$
y=\ln \left(x^{2}-1\right), 2 \leq x \leq 3
$$
## Solution

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

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

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{2}^{3} \sqrt{1+\left(\frac{2 x}{x^{2}-1}\right)^{2}} d x= \\
& =\int_{2}^{3} \sqrt{\frac{x^{4}-2 x^{2}+1+4 x^{2}}{\left(x^{2}-1\right)^{2}}} d x= \\
& =\int_{2}^{3} \sqrt{\frac{x^{4}+2 x^{2}+1}{\left(x^{2}-1\right)^{2}}} d x= \\
& =\int_{2}^{3} \sqrt{\frac{\left(x^{2}+1\right)^{2}}{\left(x^{2}-1\right)^{2}}} d x=\left|\left(x^{2}-1\right) \geq 0\right|= \\
& =\int_{2}^{3} \frac{x^{2}+1}{x^{2}-1} d x=2 \\
& =\int_{2}^{3} \frac{x^{2}-1+1+1}{x^{2}-1} d x= \\
& =\int_{2}^{3} \frac{x^{2}-1}{x^{2}-1} d x+2 \int_{2}^{3} \frac{1}{x^{2}-1} d x= \\
& =\int_{2}^{3} d x+2 \int_{2}^{3} \frac{1}{x^{2}-1} d x= \\
& =x_{2}^{3}+\left.2\left(\frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|\right)\right|_{2} ^{3}=3-2+\ln \frac{1}{2}-\ln \frac{1}{3}=1+\ln \frac{3}{2}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_17-8$ » Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

- Last edited: 10:03, 26 May 2010.
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## Problem Kuznetsov Integrals 17-9

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