## Calculus: Early Transcendentals (2nd Edition)

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