How do you integrate 1/(2v + 1)^2?
Use a $\displaystyle u$ substitution.
$\displaystyle \int{\frac{1}{(2v + 1)^2}\,dv} = \frac{1}{2}\int{2(2v+1)^{-2}\,dv}$.
Now let $\displaystyle u=2v + 1$ so that $\displaystyle \frac{du}{dv} = 2$ and the integral becomes
$\displaystyle \frac{1}{2}\int{u^{-2}\,\frac{du}{dv}\,dv}$
$\displaystyle = \frac{1}{2}\int{u^{-2}\,du}$
$\displaystyle = \frac{1}{2}\left(\frac{u^{-1}}{-1}\right) + C$
$\displaystyle = -\frac{1}{2u} + C$
$\displaystyle = -\frac{1}{2(2v+1)} + C$
$\displaystyle = -\frac{1}{4v+2} + C$.