Answer
\[ = - 3\sqrt {4 - {x^2}} + {\sin ^{ - 1}}\frac{x}{2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{3x + 1}}{{\sqrt {4 - {x^2}} }}dx} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{3x}}{{\sqrt {4 - {x^2}} }} + \frac{1}{{\sqrt {4 - {x^2}} }}} \right)dx} \hfill \\
\hfill \\
for{\text{ }}\int_{}^{} {\frac{{3x}}{{\sqrt {4 - {x^2}} }}dx} \hfill \\
\hfill \\
substituting\,\,\left( {4 - {x^2}} \right) = t\,\, \to \,xdx = - \frac{{dt}}{2} \hfill \\
\hfill \\
= - \frac{3}{2}\int_{}^{} {\frac{{dt}}{{\sqrt t }}} = - \frac{3}{2}2\sqrt t + {c_2} \hfill \\
\hfill \\
= - 3\sqrt {4 - {x^2}} + {c_2} \hfill \\
\hfill \\
for = \int_{}^{} {\frac{1}{{\sqrt {4 - {x^2}} }}} dx = {\sin ^{ - 1}}\frac{x}{2} + {c_1} \hfill \\
\hfill \\
Use\,\,\int_{}^{} {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = \,{{\sin }^{ - 1}}\frac{x}{2}\,} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
\int_{}^{} {\,\left( {\frac{{3x}}{{\sqrt {4 - {x^2}} }} + \frac{1}{{\sqrt {4 - {x^2}} }}} \right)dx} = - 3\sqrt {4 - {x^2}} + {\sin ^{ - 1}}\frac{x}{2} + C \hfill \\
\hfill \\
\end{gathered} \]