Answer
\[ = \ln \,\left( {{x^2} + 4} \right) + \frac{1}{2}{\tan ^{ - 1}}\,\left( {\frac{x}{2}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{2x + 1}}{{{x^2} + 4}}} \hfill \\
\hfill \\
split\,\,the\,\,numerator \hfill \\
\hfill \\
\frac{{2x + 1}}{{{x^2} + 4}} = \frac{{2x}}{{{x^2} + 4}} + \frac{1}{{{x^2} + 4}} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
\int_{}^{} {\frac{{2x + 1}}{{{x^2} + 4}}} = \int_{}^{} {\,\left( {\frac{{2x}}{{{x^2} + 4}} + \frac{1}{{{x^2} + 4}}} \right)dx} \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
= \ln \,\left( {{x^2} + 4} \right) + \frac{1}{2}{\tan ^{ - 1}}\,\left( {\frac{x}{2}} \right) + C \hfill \\
\hfill \\
\end{gathered} \]