Find the arc length of:

Find the arc length of r(t)=<4t^2 , lnt , -4t> from t=1 to t=a

I know the arc length is equal to the magnitude of the r'(t)
r'(t) = <8t , 1/t , -4>
but the magnitude of this comes out too messy for an integral
|r'(t)| = (64t^2 + 1/t^2 + 16)^0.5

Whats my next step?

Note that 64t^2 + 1/t^2 + 16 = (8t + 1/t)^2.

So, all you need to do is integral (8t + 1/t) from t = 1 to t = a.
==> (4t^2 + ln t) {for t = 1 to a}
= 4a^2 + ln a - 4.

I hope this helps!

This is a fairly long integration. Try looking at
http://www.wolframalpha.com/input/i=inte…
and clicking the show more steps button