Assuming that f(x) = (1/54) x^6 + 9/(16x^4):

f'(x) = (1/9)x^5 - (9/4)x^(-5)

So, √[1 + (f'(x))^2]

= √[1 + ((1/9)x^5 - (9/4)x^(-5))^2]

= √[1 + ((1/81)x^10 - 1/2 + (81/16)x^(-10))]

= √[(1/81)x^10 + 1/2 + (81/16)x^(-10)]

= √[((1/9)x^5 + (9/4)x^(-5))^2]

= (1/9)x^5 + (9/4)x^(-5)

Therefore, the arc length equals

∫(x = 1 to 2) ((1/9)x^5 + (9/4)x^(-5)) dx

= [(1/54)x^6 - (9/16)x^(-4)] {for x = 1 to 2}

= 1301/768.

I hope this helps!

L = int [√ 1 + (df/ dx)^2 ] dx

the f(x) is not clear .. use parenthesis ...