So we are told that a body of constant mass m falls vertically in air subject to forces of gravity and wind resistance the wind resistance is given by the following equation $F_r=\gamma v^2$ where $\gamma$ is a positive constant the body has no initial velocity so $v(0)=0$ we want to show that, $$v(t)=v_{\infty}(\frac{1-exp(-2\alpha t)}{1+exp(-2\alpha t)})$$
for $t\geq0$ where $v_{\infty}=-\sqrt{\frac{mg}{\gamma}}$ and $\alpha=\sqrt{\frac{\gamma g}{m}}$
So heres where iam at the actual solving isn't the problem you use newtons second to get that the total force is $F_{total}= \gamma v^2-mg$ then dividing by $m$ to get acceleration and rearranging to get $a=\frac{\gamma}{m}(v^2-\frac{mg}{\gamma})$ this is a separable ODE that can be made more simple with partial fractions. This gets me, $$\frac{1}{2}\sqrt{\frac{\gamma}{mg}}\int\frac{1}{v - \sqrt{\frac{mg}{\gamma}}}- \frac{1}{v + \sqrt{\frac{mg}{\gamma}}}dv=\int \frac{\gamma}{m}dt$$ which I evaluate and use the initial condition to get, $$ln(\frac{v+v_\infty}{v-v_\infty})=2\alpha t$$ after fiddling with this for abit the closest I can get to the answer is $$v=v_\infty\frac{1+exp(-2\alpha t)}{1-exp(-2\alpha t)}$$
Any help is much appreciated i would also like to apologize and thank in advance after writing this out i see how much of a pain it is two write out all the symbols so thanks again for the help.
You chose to work with $v<0$, which is not a good idea in general, as it can lead to sign errors. In your case, remember that $\int(dx/x)=\ln|x|$ so that:
$$ \int_0^v\left({1\over w-v_\infty}-{1\over w+v_\infty} \right)dw= \ln|v-v_\infty|-\ln|v+v_\infty|=\ln{v-v_\infty\over -v-v_\infty}. $$ This change leads to the right result.