Now

`s=intv\ dt`

[We could use s or h in this problem.]

We need to find:

`s=int[ln^2(t^3+1)](t^2)/(t^3+1)dt`

Put `u = ln(t^3+ 1)` and this gives:

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

Inside the integral, we need:

`u^2 = ln^2(t^3+ 1)`

So

`{: (s,=1/3int[u^2]du),(,=u^3/9+K),(,=(ln^3(t^3+1))/9+K) :}`

Now, since the height is `0` when `t = 0`, we substitute and obtain `K = 0`.

`s=(ln^3(t^3+1))/9`

At `t=10`, the height of the space vehicle will be:

`s=(ln^3((10)^3+1))/9=36.6\ "km"`