Determine the arclength of the curve f(x) = x^2/2 − ln x 4 for 1 ≤ x ≤ 4

Answer 1

is f(x)=x²/2-ln(4x) or f(x)=x²/2-ln(x⁴)? its hard to tell so I'm going to show the first one since it's closest to what you wrote

L=∫_a^b√(1+(f'(x))²)dx

Find f'(x)

f'(x)=x-1/(4x)(4)

f'(x)=x-1/x

L=∫₁⁴√(1+((x-1/x)²)dx

L=∫₁⁴√(1+x²-2+1/x²)dx

L=∫₁⁴√(x⁴-x²+1)/x²)dx

Since the numerator is not a perfect square I don't believe there is a way to simplify this further and integrate without a calculator. If this is a calculator question then there isn't really any need to do the simplification steps and you can integrate once you have the integral set up.

L=7.022

If it's f(x)=x²/2-ln(x⁴) I would rewrite as f(x)=x²/2-4ln(x) using log rules first then

f'(x)=x-4/x

L=∫₁⁴√(1+((x-4/x)²)dx=5.65