# What is the arclength of f(x)=ln(x+3) on x in [2,3]?

Jun 17, 2018

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{2 n - 1} \left(\frac{1}{5} ^ \left(2 n - 1\right) - \frac{1}{6} ^ \left(2 n - 1\right)\right)$ units.

#### Explanation:

$f \left(x\right) = \ln \left(x + 3\right)$

$f ' \left(x\right) = \frac{1}{x + 3}$

Arclength is given by:

$L = {\int}_{2}^{3} \sqrt{1 + \frac{1}{x + 3} ^ 2} \mathrm{dx}$

Apply the substitution $x + 3 = u$:

$L = {\int}_{5}^{6} \sqrt{1 + \frac{1}{u} ^ 2} \mathrm{du}$

For $u \in \left[5 , 6\right]$, $\frac{1}{u} ^ 2 < 1$. Take the series expansion of the square root:

$L = {\int}_{5}^{6} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{u} ^ \left(2 n\right) \mathrm{du}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{5}^{6} \mathrm{du} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{5}^{6} \frac{1}{u} ^ \left(2 n\right) \mathrm{du}$

Integrate directly:

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left[\frac{- 1}{\left(2 n - 1\right) {u}^{2 n - 1}}\right]}_{5}^{6}$

Simplify:

$L = 1 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{2 n - 1} \left(\frac{1}{5} ^ \left(2 n - 1\right) - \frac{1}{6} ^ \left(2 n - 1\right)\right)$