$\displaystyle \int \dfrac{dv}{\sqrt{1+v^2}}$
$v=\tan{t}$
$dv=\sec^2{t} \, dt$
$\displaystyle \int \dfrac{\sec^2{t}}{\sqrt{1+\tan^2{t}}} \, dt = \int \sec{t} \, dt = \ln|\sec{t}+\tan{t}|+C$
$\ln|\sqrt{1+v^2}+v|+C$
So, yes ... I'd say it's wrong.
btw ... when are you going to learn latex and start posting something legible for a change?