## Problem Statement

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

$$
y=e^{x}+e, \ln \sqrt{3} \leq x \leq \ln \sqrt{15}
$$
## 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:

$$
f^{\prime}(x)=\left(e^{x}+e\right)^{\prime}=e^{x}
$$

Then, using the above formula, we get:

$$
\begin{aligned}
L & =\int_{\ln \sqrt{3}}^{\ln \sqrt{15}} \sqrt{1+e^{2 x}} d x=\left|\begin{array}{c}
\sqrt{1+e^{2 x}}=t \\
2 x=\ln \left(t^{2}-1\right) \\
d x=\frac{t d t}{t^{2}-1} \\
t_{1}=\sqrt{3+1}=2 ; \\
t_{2}=\sqrt{15+1}=4
\end{array}\right|=\int_{2}^{4} t \cdot \frac{t d t}{t^{2}-1}= \\
& =\int_{2}^{4} \frac{t^{2}}{t^{2}-1} d t= \\
& =\int_{2}^{4} \frac{t^{2}-1+1}{t^{2}-1} d t= \\
& =\int_{2}^{4} d t+\int \frac{d t}{t^{2}-1}= \\
& =\left.t\right|_{2} ^{4}+\left.\frac{1}{2} \ln \left|\frac{t-1}{t+1}\right|\right|_{2} ^{4}=4-2+\frac{1}{2} \ln \frac{3}{5}-\frac{1}{2} \ln \frac{1}{3}=2+\frac{1}{2} \ln \left(\frac{3}{5}: \frac{1}{3}\right)=2+\frac{1}{2} \ln \frac{9}{5}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�

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$\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-30$ »

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals | Problems for Checking

- Last modified: 05:58, 26 May 2010.
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## Problem Kuznetsov Integrals 17-31

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