Finding arc length of vector valued function

[Yitin]

Homework Statement

Find arc length of the graph of
r(t) = ti + ( (t⁶/6) - (6/t⁴) )j + t√3 k
1≤t≤2

Homework Equations

Arc length = ∫ ||dr/dt|| dt
(Integral from t0 to t1 of norm of derivative of r)

The Attempt at a Solution

dr/dt = i + (t⁵ + 24/(t³) )j + √3 k

1² = 1
(√3)² = 3
(t⁵ + 24/(t³) )² = t¹⁰ + 48t² + 576/t⁶

||dr/dt|| = √(1 + 3 + t¹⁰ + 48t² + 576/t⁶)I get stuck here because I do not know how to integrate that. It doesn't seem anywhere close to a perfect square.
Did I make a mistake somewhere, or is there some kind of property to integrate this?

 

[jbunniii]

You did not differentiate the $6/t^4$ term correctly.

 

[Yitin]

Ok, so now it should be

(t⁵ +24t^-5)²
(t¹⁰ + 48 + 576t^-10)

||dr/dt|| = √( 1 + 3 + t¹⁰ + 48 + 576t^-10 )
||dr/dt|| = √( t¹⁰ + 52 + 576t^-10 )

Which I am still not sure how to integrate, but it at least looks nicer.