Question

What is the arclength of `f(x)=(x^2+24x+1)/x^2` in the interval `[1,3]`?

Answer 1

`approx 17.060431259`

Explanation:

Given `f(x)=(x^2+24x+1)/x^2`
we differentiate this function with respect to `x` using the Quotient rule

`(u/v)'=(u'v-uv')/v^2`

so we get

`f'(x)=((2x+24)x^2-(x^2+24x+1)2x)/x^4`
cancelling `xne 0`

`f'(x)=(2x^2+24x-2x^2-48x-2)/x^3`

`f'(x)=(-2(12x+1))/x^3`

so we get

`int_1^3sqrt(1+(-2(12x+1)/x^3)^2)dx`
by a numerical method we get `17.06031259`