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

Mar 16, 2016

$L = \sqrt{10} - \sqrt{2} + \ln \left(\frac{2 \sqrt{5} + \sqrt{10} - \sqrt{2} - 1}{3}\right) \cong 2.302$

#### Explanation:

To do this we need to apply the formula for the length of the curve
mentioned in:
How do you find the length of a curve using integration?

We start from
$L = {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$
$f \left(x\right) = \ln x$ => $f ' \left(x\right) = \frac{1}{x}$
Then
$L = {\int}_{1}^{3} \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx} = {\int}_{1}^{3} \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx}$

Making
$x = \tan y$
$\mathrm{dx} = {\sec}^{2} y \cdot \mathrm{dy}$
we get
$\int \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx} = \int \frac{\sec y \cdot {\sec}^{2} y}{\tan} y \mathrm{dy} = \int \left(\frac{1}{\cos} ^ 3 y\right) \left(\cos \frac{y}{\sin} y\right) \mathrm{dy} = \int \frac{\mathrm{dy}}{{\cos}^{2} y \cdot \sin y}$

We can break the integrand in partial fractions in this way
$\frac{1}{\left(1 - {\sin}^{2} x\right) \cdot \sin y} = \frac{1}{\left(1 - \sin y\right) \left(1 + \sin y\right) \sin y} = \frac{A}{1 + \sin y} + \frac{B}{1 - \sin y} + \frac{C}{\sin} y$ [1]
For $y = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}$, we get
$\frac{8}{3} = \frac{2}{3} A + 2 B + 2 C$
$- \frac{8}{3} = 2 A + \frac{2}{3} B - 2 C$ [2]
Summing the last two expressions we get
$0 = \frac{8}{3} A + \frac{8}{3} B$ => $B = - A$
Then expression [2] becomes
$- \frac{8}{3} = \frac{4}{3} A - 2 C$ => $- \frac{4}{3} = \frac{2}{3} A - C$ [3]

Since (in expression [1]) $\frac{A}{1 + \sin y} - \frac{A}{1 - \sin y} = A \cdot \frac{1 - \sin y - 1 - \sin y}{\cos} ^ 2 y = - \frac{2 A \sin y}{\cos} ^ 2 y$
the integrand can be rewritten as
$\frac{1}{{\cos}^{2} y \cdot \sin y} = - \frac{2 A \sin y}{\cos} ^ 2 y + \frac{C}{\sin} y$
For $y = \frac{\pi}{4}$
$\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^{2} \cdot \cancel{\frac{\sqrt{2}}{2}}} = - \frac{2 A \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} ^ \cancel{2} + \frac{C}{\cancel{\frac{\sqrt{2}}{2}}}$
$2 = - 2 A + C$
Summing the last expression with expression [3] we get
$\frac{2}{3} = - \frac{4}{3} A$ => $A = - \frac{1}{2}$
$\to C = 2 + 2 A = 2 - 1$ => $C = 1$

So we arrived at
$= \int \sin \frac{y}{\cos} ^ 2 y \mathrm{dy} + \int \frac{\mathrm{dy}}{\sin} y$

Making $\cos y = z$ => $\sin y \cdot \mathrm{dy} = - \mathrm{dz}$
The first part becomes
$\int - \frac{\mathrm{dz}}{z} ^ 2 = \frac{1}{z} = \frac{1}{\cos} y$

Therefore
$= \frac{1}{\cos} y + \ln | \csc y - \cot y | + c o n s t .$
But
$x = \tan y$ => $\sin y = x \cos y$
${\sin}^{2} y + {\cos}^{2} y = 1$ => $\left({x}^{2} + 1\right) {\cos}^{2} y = 1$ => $\cos y = \frac{1}{\sqrt{{x}^{2} + 1}}$
$\to \sin y = x . \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right)$ => $\sin y = \frac{x}{\sqrt{{x}^{2} + 1}}$
So
$= \sqrt{{x}^{2} + 1} + \ln | \frac{\sqrt{{x}^{2} + 1}}{x} - \frac{1}{x} | + c o n s t .$
$= \sqrt{{x}^{2} + 1} + \ln | \frac{\sqrt{{x}^{2} + 1} - 1}{x} | + c o n s t .$

Finally
$L = \left(\sqrt{{x}^{2} + 1} + \ln | \frac{\sqrt{{x}^{2} + 1} - 1}{x} |\right) {|}_{1}^{3}$
$L = \sqrt{10} + \ln \left(\frac{\sqrt{10} - 1}{3}\right) - \sqrt{2} - \ln \left(\sqrt{2} - 1\right)$
$L = \sqrt{10} - \sqrt{2} + \ln \left(\frac{\sqrt{10} - 1}{3 \left(\sqrt{2} - 1\right)}\right)$
But
$\frac{\sqrt{10} - 1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = 2 \sqrt{5} + \sqrt{10} - \sqrt{2} - 1$
So

$L = \sqrt{10} - \sqrt{2} + \ln \left(\frac{2 \sqrt{5} + \sqrt{10} - \sqrt{2} - 1}{3}\right)$