Answer
$$\frac{3}{{\sqrt {{x^2} + 3} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - 3x}}{{{{\left( {{x^2} + 3} \right)}^{3/2}}}}} dx \cr
& {\text{Integrate by substitution }} \cr
& {\text{Let }}u = {x^2} + 3,{\text{ }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}} \cr
& {\text{Substitute}} \cr
& \int {\frac{{ - 3x}}{{{{\left( {{x^2} + 3} \right)}^{3/2}}}}} dx = \int {\frac{{ - 3x}}{{{u^{3/2}}}}} \left( {\frac{{du}}{{2x}}} \right) \cr
& = - \frac{3}{2}\int {\frac{{du}}{{{u^{3/2}}}}} \cr
& = - \frac{3}{2}\int {{u^{ - 3/2}}du} \cr
& {\text{By the power rule}} \cr
& = - \frac{3}{2}\left( {\frac{{{u^{ - 1/2}}}}{{ - 1/2}}} \right) + C \cr
& = 3{u^{ - 1/2}} + C \cr
& = \frac{3}{{\sqrt u }} + C \cr
& {\text{Substitute back }}u = {x^2} + 3 \cr
& = \frac{3}{{\sqrt {{x^2} + 3} }} + C \cr} $$
