Integration : Find the Arc Length over an Interval

Find the arc length of the graph of the function over the indicated interval:

y=1/6 x^3 + 1/(2x^2), [1,3]

I know S = Intergral( sqr( 1 + [f'(x)]^2 )) dx from 1 to 3.

I get y’ = [ 1/9 x^4 – 1/3 + 1/(4x^4) ] dx

Therefore,

S = Intergal( sqr( 1 + 1/9 x^4 – 1/3 + 1/(4x^4) )) dx from 1 to 3

= Intergal( sqr( 2/3 + 1/9 x^4 + 1/(4x^4) )) dx from 1 to 3

= Intergal( sqr( 2/3 ) + 1/3 x^2 + 1/(2x^2) ) dx from 1 to 3

Taking the Intergral:

S = [ sqr(2/3) x + 1/9 x^3 + ???.