I figured it out by multiplying 1/3 to 6 from f(x)dx = 6. I don't know why I needed to do that though.
u = x^2
du = 2x dx
(1/2)du = x dx
Now, substitute u for x^2 and (1/2)du for x dx. Change your bounds in terms of u (which conveniently become the bounds in the first integral). Your integral is now identical to the first integral with the exception of it being in terms of u and the 1/2 constant out front. Your answer then becomes (1/2) * 6 = 3.