Answer
$$\int\frac{x^2+6x}{(x^2+3)^2}dx=\frac{1}{2\sqrt{3}}\tan^{-1}\frac{x}{\sqrt3}-\frac{x+6}{2(x^2+3)}+C$$
Work Step by Step
$$A=\int\frac{x^2+6x}{(x^2+3)^2}dx$$ $$A=\int\frac{x^2+3}{(x^2+3)^2}dx-\int\frac{3}{(x^2+3)^2}dx+\int\frac{6x}{(x^2+3)^2}dx$$ $$A=\int\frac{dx}{x^2+3}-3\int\frac{dx}{(x^2+3)^2}+3\int\frac{2x}{(x^2+3)^2}dx$$
For the first integral, use Formula 32, which states that
$$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$
for $a=\sqrt3$
For the second integral, use Formula 33, which states that
$$\int\frac{dx}{(a^2+x^2)^2}=\frac{x}{2a^2(a^2+x^2)}+\frac{1}{2a^3}\tan^{-1}\frac{x}{a}+C$$
for $a=\sqrt3$
For the final integral, take $u=x^2+3$, which makes $du=2xdx$
$$A=\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-3\Big[\frac{x}{2\times3(x^2+3)}+\frac{1}{2\times3\sqrt3}\tan^{-1}\frac{x}{\sqrt3}\Big]+3\int\frac{du}{u^2}$$ $$A=\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-3\Big[\frac{x}{6(x^2+3)}+\frac{1}{6\sqrt3}\tan^{-1}\frac{x}{\sqrt3}\Big]-\frac{3}{u}+C$$ $$A=\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-\frac{x}{2(x^2+3)}-\frac{1}{2\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-\frac{3}{x^2+3}+C$$ $$A=\Big(\frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-\frac{1}{2\sqrt3}\tan^{-1}\frac{x}{\sqrt3}\Big)-\Big(\frac{x}{2(x^2+3)}+\frac{3}{x^2+3}\Big)+C$$ $$A=\frac{1}{2\sqrt{3}}\tan^{-1}\frac{x}{\sqrt3}-\frac{x+6}{2(x^2+3)}+C$$
