Answer
$$\,\,\int {\frac{{3 - {x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}}} dx = \frac{x}{{{x^2} + 3}} + C$$
Work Step by Step
$$\eqalign{
& {\text{Let }}\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 3}}} \right] \cr
& {\text{using the quotient rule}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 3}}} \right] = \frac{{\left( {{x^2} + 3} \right)\left( x \right)' - \left( x \right)\left( {{x^2} + 3} \right)'}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Differentiating}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 3}}} \right] = \frac{{\left( {{x^2} + 3} \right)\left( 1 \right) - \left( x \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 3}}} \right] = \frac{{{x^2} + 3 - 2{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 3}}} \right] = \frac{{3 - {x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Thus}}{\text{,}}\,\,\,\,{\text{a corresponding integration formula is}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\int {\frac{{3 - {x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}}} dx = \frac{x}{{{x^2} + 3}} + C \cr} $$
