I tried solving this equation using several different methods but I never got to an answer. I tried using partial integration and at last I got $m/\sqrt{mg}\ln mg-kv^2 = t+c$
I even tried using substitution but I could not get an answer. I would be really grateful if someone could help me completing my solution or even linking me to somewhere where there is a conclusive answer. Thanks a lot!
Since $dt=\frac{m dv}{mg-kv^2}$, you can easily express $t-t_0$ (say) in terms of $v$ using the hyperbolic arctangent. You can then invert this to get $v$ in terms of $t-t_0$. I'll leave integration to $x=x_0+\int_0^t v(t^\prime) dt^\prime$ to you, using the fact that $\int\tanh u du=\ln\cosh u+C$.