## Thomas' Calculus 13th Edition

$\dfrac{1}{2\sqrt{3}}\tan^{-1} (\dfrac{x}{\sqrt3}) -\dfrac{x+6}{2(x^2+3)}+C$
Let $I=\int\dfrac{x^2+6x}{(x^2+3)^2}dx=\int\dfrac{dx}{x^2+3}-3\int\dfrac{dx}{(x^2+3)^2}+3\int\dfrac{2x}{(x^2+3)^2} \space dx$ Suppose $u=x^2+3\implies du=2xdx$ $I=\dfrac{1}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3}-3\Big[\frac{x}{2\times3(x^2+3)}+\dfrac{1}{2\times3\sqrt3}\tan^{-1}\frac{x}{\sqrt3}\Big]+3 \times \int\dfrac{du}{u^2} \\=\dfrac{1}{\sqrt3} \space \tan^{-1}(\dfrac{x}{\sqrt3})-3 [\dfrac{x}{6(x^2+3)}+\dfrac{1}{6\sqrt3} \space \tan^{-1} (\dfrac{x}{\sqrt3}) ]-\dfrac{3}{u}+C \\=\dfrac{1}{\sqrt3}\tan^{-1}(\dfrac{x}{\sqrt3})-\dfrac{x}{2(x^2+3)}-\dfrac{1}{2\sqrt3}\tan^{-1} (\dfrac{x}{\sqrt3}) -\dfrac{3}{x^2+3}+C \\=\dfrac{1}{2\sqrt{3}}\tan^{-1} (\dfrac{x}{\sqrt3}) -\dfrac{x+6}{2(x^2+3)}+C$