# What is the arc length of f(x)=lnx  in the interval [1,5]?

Mar 7, 2018

The arc length is $\sqrt{26} - \sqrt{2} + \ln 5 - \ln \left(\frac{1 + \sqrt{26}}{1 + \sqrt{2}}\right)$ units.

#### Explanation:

$f \left(x\right) = \ln x$

$f ' \left(x\right) = \frac{1}{x}$

Arc length is given by:

$L = {\int}_{1}^{5} \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx}$

Rearrange:

$L = {\int}_{1}^{5} \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx}$

Apply the substitution $x = \tan \theta$:

$L = \int \sec \frac{\theta}{\tan} \theta \cdot {\sec}^{2} \theta d \theta$

Rewrite as:

$L = \int \csc \theta \cdot \left({\tan}^{2} \theta + 1\right) d \theta$

Hence

$L = \int \left(\sec \theta \tan \theta + \csc \theta\right) d \theta$

Integrate term by term:

$L = \left[\sec \theta - \ln | \csc \theta + \cot \theta |\right]$

Reverse the substitution:

$L = {\left[\sqrt{1 + {x}^{2}} - \ln | \frac{1 + \sqrt{1 + {x}^{2}}}{x} |\right]}_{1}^{5}$

Insert the limits of integration:

$L = \sqrt{26} - \sqrt{2} + \ln 5 - \ln \left(\frac{1 + \sqrt{26}}{1 + \sqrt{2}}\right)$