# What is the arc length of f(x) = ln(x)  on x in [1,3] ?

The arc length $\textcolor{red}{s = 2.3019875345832} \text{ }$units

#### Explanation:

Given function $f \left(x\right) = y = \ln x$ with x from 1 to 3

The formula for arc length $s$

$s = {\int}_{1}^{3} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{x}\right) \frac{\mathrm{dx}}{\mathrm{dx}} = \frac{1}{x}$

$s = {\int}_{1}^{3} \sqrt{1 + {\left(\frac{1}{x}\right)}^{2}} \mathrm{dx}$

$s = {\int}_{1}^{3} \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx}$

Using Algebraic Substitution

Let $z = \sqrt{{x}^{2} + 1}$
and ${z}^{2} = {x}^{2} + 1$
Differentiating both sides
$2 z \cdot \mathrm{dz} = 2 x \cdot \mathrm{dx} + 0$
$2 z \cdot \mathrm{dz} = 2 x \cdot \mathrm{dx}$

dividing both sides by ${x}^{2}$
$2 z \cdot \mathrm{dz} = 2 x \cdot \mathrm{dx}$

$\frac{2 z}{x} ^ 2 \cdot \mathrm{dz} = \frac{2 x}{x} ^ 2 \cdot \mathrm{dx} = \frac{2 \mathrm{dx}}{x}$

$\frac{2 z}{x} ^ 2 \cdot \mathrm{dz} = \frac{2 \mathrm{dx}}{x}$

$\frac{2 z \cdot \mathrm{dz}}{{z}^{2} - 1} = \frac{2 \mathrm{dx}}{x}$

and
$\frac{z \cdot \mathrm{dz}}{{z}^{2} - 1} = \frac{\mathrm{dx}}{x}$

Let us go back to

$s = {\int}_{1}^{3} \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx} = {\int}_{1}^{3} \left(\sqrt{{x}^{2} + 1}\right) \left(\frac{\mathrm{dx}}{x}\right)$

Replacing now with variable $z$

$s = {\int}_{1}^{3} \left(\sqrt{{x}^{2} + 1}\right) \left(\frac{\mathrm{dx}}{x}\right) = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(z\right) \left(\frac{z \cdot \mathrm{dz}}{{z}^{2} - 1}\right)$

Because $z = \sqrt{{x}^{2} + 1} \text{ }$
If $x$ is from $x = 1$ to $x = 3$, then, $z$ is from $z = \sqrt{2}$ to $z = \sqrt{10}$

Now we integrate

$s = {\int}_{1}^{3} \left(\sqrt{{x}^{2} + 1}\right) \left(\frac{\mathrm{dx}}{x}\right) = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(z\right) \left(\frac{z \cdot \mathrm{dz}}{{z}^{2} - 1}\right)$

$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(z\right) \left(\frac{z \cdot \mathrm{dz}}{{z}^{2} - 1}\right)$
$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(\frac{{z}^{2} \cdot \mathrm{dz}}{{z}^{2} - 1}\right)$
$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \frac{\left({z}^{2} - 1 + 1\right) \cdot \mathrm{dz}}{{z}^{2} - 1}$

$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(\frac{{z}^{2} - 1}{{z}^{2} - 1} + \frac{1}{{z}^{2} - 1}\right) \cdot \mathrm{dz}$

$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(1 + \frac{1}{{z}^{2} - 1}\right) \cdot \mathrm{dz}$

$s = {\int}_{\sqrt{2}}^{\sqrt{10}} \left(1 + \frac{1}{{z}^{2} - 1}\right) \cdot \mathrm{dz} = {\left[z + \frac{1}{2} \cdot \ln \left(\frac{z - 1}{z + 1}\right)\right]}_{\sqrt{2}}^{\sqrt{10}}$

$s = \sqrt{10} + \frac{1}{2} \ln \left(\frac{\sqrt{10} - 1}{\sqrt{10} + 1}\right) - \left(\sqrt{2} + \frac{1}{2} \ln \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right)\right)$

$\textcolor{red}{s = 2.3019875345832} \text{ }$units

God bless....I hope the explanation is useful.