# What is the arclength of f(x)=x/(x-5) in [0,3]?

Jun 26, 2018

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{8}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\cot}^{- 1} \left(\sqrt{5}\right) + {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} / \left(n - m\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \sin \left(\left(2 n - 2 m\right) {\cot}^{- 1} \left(\sqrt{5}\right)\right) \cos \left(\left(2 n - 2 m\right) {\tan}^{- 1} \left(\frac{4}{\sqrt{5}}\right)\right)\right\}$

#### Explanation:

$f \left(x\right) = \frac{x}{x - 5}$, $x \in \left[0 , 3\right]$

Let $y = f \left(x\right)$. We can invert:

$x = \frac{5}{y - 1} + 5$, $y \in \left[- \frac{3}{2} , 0\right]$

Take the derivative with respect to $y$:

$x ' = - \frac{5}{y - 1} ^ 2$

Arclength is given by:

$L = {\int}_{- \frac{3}{2}}^{0} \sqrt{1 + \frac{25}{y - 1} ^ 4} \mathrm{dy}$

Apply the substitution $1 - y = \frac{u}{2}$:

$L = \frac{1}{2} {\int}_{2}^{5} \sqrt{1 + \frac{400}{u} ^ 4} \mathrm{du}$

Complete the square:

$L = \frac{1}{2} {\int}_{2}^{5} \sqrt{{\left(1 + \frac{20}{u} ^ 2\right)}^{2} - \frac{40}{u} ^ 2} \mathrm{du}$

Factor out the larger piece:

$L = \frac{1}{2} {\int}_{2}^{5} \left(1 + \frac{20}{u} ^ 2\right) \sqrt{1 - \frac{40 {u}^{2}}{{u}^{2} + 20} ^ 2} \mathrm{du}$

For $u \in \left[2 , 5\right]$, $\frac{40 {u}^{2}}{{u}^{2} + 20} ^ 2 < 1$. Take the series expansion of the square root:

$L = \frac{1}{2} {\int}_{2}^{5} \left(1 + \frac{20}{u} ^ 2\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{40 {u}^{2}}{{u}^{2} + 20} ^ 2\right)}^{n}\right\} \mathrm{du}$

Isolate the $n = 0$ term and simplify:

$L = \frac{1}{2} {\int}_{2}^{5} \left(1 + \frac{20}{u} ^ 2\right) \mathrm{du} + \frac{1}{2} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 40\right)}^{n} {\int}_{2}^{5} \frac{1}{u} {\left(\frac{u}{{u}^{2} + 20}\right)}^{2 n - 1} \mathrm{du}$

Apply the substitution $u = \sqrt{20} \tan \theta$:

$L = \frac{1}{2} {\left[u - \frac{20}{u}\right]}_{2}^{5} + \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} \int {\left(\frac{\tan \theta}{{\sec}^{2} \theta}\right)}^{2 n - 2} d \theta$

Isolate the $n = 1$ term and simplify:

$L = \frac{9}{2} - \sqrt{5} \int d \theta + \sqrt{5} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} \int {\left(\frac{\tan \theta}{{\sec}^{2} \theta}\right)}^{2 n - 2} d \theta$

Apply the appropriate half-angle Trigonometric identities:

$L = \frac{9}{2} - \sqrt{5} {\left[{\tan}^{- 1} \left(\frac{u}{\sqrt{20}}\right)\right]}_{2}^{5} + 4 \sqrt{5} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\frac{- 1}{2}\right)}^{n} \int {\sin}^{2 n - 2} \left(2 \theta\right) d \theta$

Rescale $n$:

$L = \frac{9}{2} - \sqrt{5} \left({\tan}^{- 1} \left(\frac{\sqrt{5}}{2}\right) - {\tan}^{- 1} \left(\frac{1}{\sqrt{5}}\right)\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{2}\right)}^{n} \int {\sin}^{2 n} \left(2 \theta\right) d \theta$

Apply the appropriate Trigonometric power-reduction formula:

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{2}\right)}^{n} \int \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} \left(\begin{matrix}2 n \\ m\end{matrix}\right) \cos \left(\left(4 n - 4 m\right) \theta\right)\right\} d \theta$

Integrate term by term:

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{8}\right)}^{n} \left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) \theta + \frac{1}{2} {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} / \left(n - m\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \sin \left(\left(4 n - 4 m\right) \theta\right)\right]$

Reverse the last substitution:

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{8}\right)}^{n} {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\tan}^{- 1} \left(\frac{u}{\sqrt{20}}\right) + \frac{1}{2} {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} / \left(n - m\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \sin \left(\left(4 n - 4 m\right) {\tan}^{- 1} \left(\frac{u}{\sqrt{20}}\right)\right)\right]}_{2}^{5}$

Insert the limits of integration:

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{8}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\cot}^{- 1} \left(\sqrt{5}\right) + \frac{1}{2} {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} / \left(n - m\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \left(\sin \left(\left(4 n - 4 m\right) {\tan}^{- 1} \left(\frac{\sqrt{5}}{2}\right)\right) - \sin \left(\left(4 n - 4 m\right) {\tan}^{- 1} \left(\frac{1}{\sqrt{5}}\right)\right)\right)\right\}$

Apply the appropriate Trigonometric angle-addition identities:

$L = \frac{9}{2} - \sqrt{5} {\cot}^{- 1} \left(\sqrt{5}\right) - 2 \sqrt{5} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{1}{8}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\cot}^{- 1} \left(\sqrt{5}\right) + {\sum}_{m = 0}^{n - 1} {\left(- 1\right)}^{n - m} / \left(n - m\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) \sin \left(\left(2 n - 2 m\right) {\cot}^{- 1} \left(\sqrt{5}\right)\right) \cos \left(\left(2 n - 2 m\right) {\tan}^{- 1} \left(\frac{4}{\sqrt{5}}\right)\right)\right\}$