problem with integral

$$$\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}\frac{\sqrt{3}}{2}$$ $

I assume that this is meant to be

$$\frac{\sqrt{3}}{2}\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}dv\\\\ =\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}v^2+\frac{3}{4}}\;dv\\\\ =\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}(v^2+1)}\;dv\\\\ =\frac{\sqrt{3}\times 4}{2\times 3}\int \frac{1}{(v^2+1)}\;dv\\\\ =\frac{2}{\sqrt 3}\int \frac{1}{1+v^2}\;dv\\\\ =\frac{2}{\sqrt 3}\;tan^{-1}\;v +c$$

I think that is correct.