Evaluate the integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x-3}{(x^2+2x+4{)}^2}dx$

Select the correct answer.

$\frac{-1}{2(x^2+2x+4)}-\frac{2\sqrt[\phantom{2}]{3}}{9}ta{n}^{-1}(\frac{x+1}{\sqrt[\phantom{2}]{3}})-\frac{2(x+1)}{3(x^2+2x+4)}+c$

$\frac{-1}{2(x^2+2x+4)}-\frac{2\sqrt[\phantom{2}]{3}}{9}ta{n}^{-1}(\frac{x+1}{\sqrt[\phantom{2}]{3}})-\frac{4(x+1)}{3(x^2+2x+4)}+c$

$\frac{-1}{2(x^2+2x+4)}-\frac{4\sqrt[\phantom{2}]{3}}{9}ta{n}^{-1}(\frac{x+1}{\sqrt[\phantom{2}]{3}})-\frac{4(x+1)}{3(x^2+2x+4)}+c$

$\frac{-1}{x^2+2x+4}-\frac{4\sqrt[\phantom{2}]{3}}{9}ta{n}^{-1}(\frac{x+1}{\sqrt[\phantom{2}]{3}})-\frac{2(x+1)}{3(x^2+2x+4)}+c$