Answer
\[ = - \frac{{{x^3}}}{3} + 4x - 5{\tan ^{ - 1}}\frac{x}{2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{6 - {x^4}}}{{{x^2} + 4}}dx} \hfill \\
\hfill \\
{\text{using}}\,\,{\text{the}}\,\,{\text{long}}\,\,{\text{division}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( { - {x^2} + 4 - \frac{{10}}{{{x^2} + 4}}} \right)dx} \hfill \\
\hfill \\
Usi\,ng\,\int_{}^{} {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}\,{{\tan }^{ - 1}}\frac{x}{a}} \,\,for\,the\,last\,term \hfill \\
\hfill \\
= - \frac{{{x^3}}}{3} + 4x - 10\frac{1}{2}{\tan ^{ - 1}}\frac{x}{2} + C \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= - \frac{{{x^3}}}{3} + 4x - 5{\tan ^{ - 1}}\frac{x}{2} + C \hfill \\
\end{gathered} \]
