Q:

Seemingly impossible square root solution

How is this equation possible? Maybe its the fact that I'm $20$ hours into a $40$ hour cram session, but this just doesn't make sense to me.
It's a standard Arc Length equation, $c(t)= (4-2t)\hat{\textbf{i}} +\ln(4t)\hat{\textbf{j}}+(5+t^2)\hat{\textbf{k}}$
I'm down to the point of finding the magnitude of $C'$ for the integral, but I'm stuck at this point
$(4+(\frac{1}{t^2})+4t^2)^{\frac{1}{2}}$ My solution guide says this is equal to $\frac{2t^2 + 1}{t}$ but I just don't see it, please help. 

A:

$$ 4 + \frac{1}{t^2} + 4t^2 = \frac{4t^2 + 1 + 4t^4}{t^2} = \frac{(2t^2 + 1)^2}{t^2} = \left(\frac{2t^2 + 1}{t}\right)^2$$
So the square root is as given.

