Kinematics Using Integral Calculus

The problem is stated as follows: The acceleration of a rocket travelling upward is given by $\displaystyle a=(6+0.02s) m/s$, where s is in meters. Determine the time needed for the rocket to reach an altitude of s=100m. Initially, v=0, s=0, t=0.

$\displaystyle vdv=ads$

$\displaystyle \int{_{v_0} ^v}vdv=\int{_{s_0} ^s}(6+0.02s)ds$

$\displaystyle \frac{1}{2}v^2=6s+0.01s^2$

$\displaystyle v=\sqrt{12s+0.02s^2}$

Then, we have:

$\displaystyle dt=ds/v$

$\displaystyle \int dt=\int ds/v$

$\displaystyle t=\int ds[12s+0.02s^2]^{-1}$

and it is at this point my mind draws a blank.... I am taking this course at the same time as integral calculus math is being tought to me, and so I really have not much to go off of for this. Any suggestions?