Answer
\[ - 2\ln \,\left( {{x^2} + 3} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{ - 4x}}{{{x^2} + 3}}dx} \hfill \\
Let\,\,\,u = {x^2} + 3\,\,so\,\,that \hfill \\
\,\,\,du = 2xdx \hfill \\
\int_{}^{} {\frac{{ - 4x}}{{{x^2} + 3}}dx = - 2\int_{}^{} {\frac{{2x}}{{{x^2} + 3}}dx} } \hfill \\
= - 2\int_{}^{} {\frac{{du}}{u}} \hfill \\
Integrating \hfill \\
- 2\ln \left| u \right| + C \hfill \\
Substituting\,\,u = {x^2} + 3\,\,for\,\,u\,\,gives \hfill \\
- 2\ln \,\left( {{x^2} + 3} \right) + C \hfill \\
\end{gathered} \]
