`t=int(dv)/(20-v)`

Let `u=20-v`, then `du=-dv`

So `t=int(dv)/(20-v)`

`=-int(du)/u`

`=-ln|u|+K`

`=-ln|20-v|+K`

When `t=0`, `v=0`

So `0=-ln\ 20+K`

Therefore `K=ln\ 20`

So `t=ln\ 20-ln(20-v)`

Applying the log laws, we get:

`t=ln(20/(20-v))`

Taking "`e` to both sides":

`e^t=20/(20-v)`

Inverting the fraction gives:

`e^-t=(20-v)/20`

`20e^-t=20-v`

` v=20-20e^-t`

`v=20(1-e^-t)`