You need to use the following formula to evaluate the given arclength such that:

`l = int_a^b sqrt(1 + (f'(x))^2)dx`

Reasoning by analogy yields:

`int_3^8 sqrt(1 + ((1/8(-2x^2+4ln(x)))')^2)dx`

You need to evaluate the derivative of the function `f(x) = 1/8(-2x^2+4ln(x))` , using quotient rule, such that:

`f'(x) = (-(-16x^2+32lnx)')/(64(-2x^2+4lnx)^2)`

`f'(x) = (32x- 32/x)/(64(-2x^2+4lnx)^2)`

Factoring out 32 yields:

`f'(x) = (x - 1/x)/(2(-2x^2+4lnx)^2)`

Raising to square yields:

`f'(x)^2 = (x - 1/x)^2/(4(-2x^2+4lnx)^4)`

You need to evaluate the integral`int_3^8 sqrt(1 + ((1/8(-2x^2+4ln(x)))')^2)dx` such that:

`int sqrt(1 + (x - 1/x)^2/(4(-2x^2+4lnx)^4))dx`

`int (sqrt (4(-2x^2+4lnx)^4 + (x - 1/x)^2))/(2(-2x^2+4lnx)^2)dx`

**Hence, evaluating the arclength of the given curve yields `l =int_3^8 (sqrt (4(-2x^2+4lnx)^4 + (x - 1/x)^2))/(2(-2x^2+4lnx)^2)dx.` **