# What is the arclength of f(x)=(x-2)/(x^2-x-2) on x in [1,2]?

May 16, 2018

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 n - 1} \left(\frac{1}{2} ^ \left(4 n - 1\right) - \frac{1}{3} ^ \left(4 n - 1\right)\right)$

#### Explanation:

$f \left(x\right) = \frac{x - 2}{{x}^{2} - x - 2} = \frac{1}{x + 1}$

$f ' \left(x\right) = - \frac{1}{x + 1} ^ 2$

Arclength is given by:

$L = {\int}_{1}^{2} \sqrt{1 + \frac{1}{x + 1} ^ 4} \mathrm{dx}$

For $x \in \left[1 , 2\right]$, $\frac{1}{x + 1} ^ 4 < 1$. Take the series expansion of the square root:

$L = {\int}_{1}^{2} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{x + 1} ^ \left(4 n\right) \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{1}^{2} \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{1}^{2} \frac{1}{x + 1} ^ \left(4 n\right) \mathrm{dx}$

Integrate directly:

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{1 - 4 n} {\left[\frac{1}{x + 1} ^ \left(4 n - 1\right)\right]}_{1}^{2}$

Insert the limits of integration:

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 n - 1} \left(\frac{1}{2} ^ \left(4 n - 1\right) - \frac{1}{3} ^ \left(4 n - 1\right)\right)$