To find the displacement, we need to evaluate:

`int_2^3(t^2+1)/((t^3+3t)^2)dt`

Put `u=t^3+3t`, then `du=(3t^2+3)dt=3(t^2+1)dt`

So `(du)/3=(t^2+1)dt`

So we have:

`int_2^3(t^2+1)/((t^3+3t)^2)=1/3int_(t=2)^(t=3)1/u^2du`

`=1/3int_(t=2)^(t=3)u^-2du`

`=-1/3[1/u]_(t=2)^(t=3)`

`=-1/3[1/(t^3+3t)]_2^3`

`=-1/3[1/(3^3+3(3))-1/(2^3+3(2))]`

`=-1/3[1/36-1/14]`

`=0.014550`

So the displacement of the object from time `t=2` to `t=3` is `0.015` units.