# What is the arc length of f(x)= 1/(2+x)  on x in [1,2] ?

May 22, 2018

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

#### Explanation:

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

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

Arc length is given by:

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

Apply the substitution $2 + x = u$:

$L = {\int}_{3}^{4} \sqrt{1 + \frac{1}{u} ^ 4} \mathrm{du}$

For $u \in \left[3 , 4\right]$, $\frac{1}{u} ^ 4 < 1$. Take the series expansion of the square root:

$L = {\int}_{3}^{4} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{u} ^ \left(4 n\right) \mathrm{du}$

Isolate the $n = 0$ term and simplify:

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

The remaining integrals are trivial:

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

Simplify:

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