Difficult Integration Question Need Help

Hi I am really stuck on this, it is a calculator problem but I can't get the answer of 5.933. Please show help with steps :
Object traveling in straight line has position s(t) at time t. If initial position is s(0) = 3 and the velocity of the object is v(t) = cube root of (1+2t^2) , what is the position at time = 2 ?
I know this is integration and it was an ugly result on my calculator but I couldn't find an answer regardless

The integral from zero to two of your velocity function is 2.933. Combine with initial position of 3 to get final position 5.933. But the integral is a messy hypergeometric function.

I take it the problem is to evaluate the integral numerically without trying to do anything analytically, correct? I have no idea what kind of calculator you have or what kind of functions you would be capable of entering into the calculator. There are calculators available on the web to do different methods of numerical integrations.

I used Simpson's rule and got the integral from 0 to 2 of the function (1+2t^2)^(1/3) to be 2.933. Then you have to add 3 to it since you are told the position at t=0 is s(0)=3 and that gives you the required answer.

Well first, find the antiderivative of your v(t) which will give you s(t) then and add a 3 since your c=3.
Then just plug 2 into your new (s(t)). It's just finding antiderivatives and that should give you your answer.
So it your final equation should be the antiderivative of v(t) +3, and plug in two
It's not a true integration since it doesn't ask how much has changed from one time to the next

ds/dt=v(t)

ds=v(t)dt
s=integral (cube root of (1+2t^2) dt )
if t=0 s(0)=3 (to find the constant value of the integral.

then calculate s(2)=?