Answer
$$\frac{3}{2}{\tan ^{ - 1}}\frac{v}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{3}{{4 + {v^2}}}} dv \cr
& {\text{take out the constant}} \cr
& = 3\int {\frac{1}{{4 + {v^2}}}} dv \cr
& {\text{from the table 4}}{\text{.10 }}\int {\frac{{dx}}{{{a^2} + {x^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + C \cr
& {\text{letting }}a = 2,{\text{ }}v = x \cr
& = 3\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{v}{2}} \right) + C \cr
& = \frac{3}{2}{\tan ^{ - 1}}\frac{v}{2} + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dv}}\left( {\frac{3}{2}{{\tan }^{ - 1}}\frac{v}{2} + C} \right) \cr
& {\text{ = }}\frac{{\text{3}}}{2}\frac{d}{{dv}}\left( {{{\tan }^{ - 1}}\frac{v}{2}} \right) + \frac{d}{{dv}}\left( C \right) \cr
& {\text{ = }}\frac{{\text{3}}}{2}\left( {\frac{2}{{{{\left( 2 \right)}^2} + {v^2}}}} \right) + 0 \cr
& = \frac{3}{{4 + {v^2}}} \cr} $$
