Now

`s=int ("velocity") dt`

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

We need to find:

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

We put the following (since any other possible substitution doesn't actually work):

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

We need to find `(du)/(dt)`. We have a function of a function, so we'll need to use the Chain Rule to find that derivative. We substitute a new variable, `v`:

`v=t^3+1`, giving `u = ln v`.

Now `(dv)/(dt) = 3t^2` and `(du)/(dv) = 1/v = 1/(t^3+1)`.

Using the Chain Rule, we get:

`(du)/(dt)=(du)/(dv)(dv)/(dt)`

` = 1/(t^3+1) xx (3t^2)`

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

Writing this as a differential in the form we need for the integral, we have:

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

Going back to the original integral with the substitution `u = ln(t^3+ 1)`, we get:

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

So

`s=1/3int[u^2]du`

`=1/3 xx u^3/3+K`

`=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"`