# What is the arc length of f(x)=2/x^4-1/(x^3+7)^6 on x in [3,oo]?

Jun 18, 2018

$\infty$

#### Explanation:

The arc length formula is:
$s = {\int}_{{x}_{0}}^{x + 1} \sqrt{1 + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

In this case, $f \left(x\right) = \frac{2}{x} ^ 4 - \frac{1}{{x}^{3} + 7} ^ 6$, which immediately suggests that the resulting mess will be too complex to evaluate. Let's produce the mess anyhow:
$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{8}{x} ^ 5 + \frac{18 {x}^{2}}{{x}^{3} + 7} ^ 7$, so
$s = {\int}_{3}^{\infty} \sqrt{1 + \frac{64}{x} ^ 10 - \frac{288}{{x}^{3} {\left({x}^{3} + 7\right)}^{7}} + \frac{324 {x}^{4}}{{x}^{3} + 7} ^ 14} \mathrm{dx}$

Yep, that is way too messy to hope to tackle. Feeding it in to integrals.wolfram.com doesn't even produce an answer.

However, the rather strange infinite limit on the requested arc length interval tells us the answer straight away. Whatever the function $f \left(x\right)$ does (and it exists for all values in the interval), you're measuring the length of a curve running from a finite value to an infinite one, so the length of the curve cannot be less than infinity - if that concept can be said to have meaning in this context at all. Thus, immediately from inspection: $s = \infty$.