## Calculus: Early Transcendentals (2nd Edition)

$\begin{gathered} = 2{\sin ^{ - 1}}x + 3\sqrt {1 - {x^2}} + c \hfill \\ \hfill \\ \end{gathered}$
$\begin{gathered} \int_{}^{} {\frac{{2 - 3x}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \hfill \\ {\text{split}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{2}{{\sqrt {1 - {x^2}} }} - \frac{{3x}}{{\sqrt {1 - {x^2}} }}} \right)dx} \hfill \\ \hfill \\ for = \int_{}^{} {\frac{2}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \hfill \\ integrate{\text{ use }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x + C} \hfill \\ \hfill \\ \int_{}^{} {\frac{2}{{\sqrt {1 - {x^2}} }}dx} = 2{\sin ^{ - 1}}x + C \hfill \\ \hfill \\ for\,\,\, - \int_{}^{} {\frac{{3x}}{{\sqrt {1 - {x^2}} }}} dx \hfill \\ \hfill \\ set\,\,\,\left( {1 - {x^2}} \right) = t\,\, \to \,\,\,t = xdx = - \frac{{dt}}{2} \hfill \\ \hfill \\ - \int_{}^{} {\frac{{3x}}{{\sqrt {1 - {x^2}} }}} dx = \frac{3}{2}\int_{}^{} {\frac{{dt}}{{\sqrt t }}} \hfill \\ \hfill \\ in\,tegrate \hfill \\ \hfill \\ = \frac{3}{2}2\sqrt t + {c_2} \hfill \\ \hfill \\ substituting\,back\,\,t = \sqrt {1 - {x^2}} \hfill \\ \hfill \\ = 3\sqrt {1 - {x^2}} + {c_2} \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{2}{{\sqrt {1 - {x^2}} }} - \frac{{3x}}{{\sqrt {1 - {x^2}} }}} \right)dx} = 2{\sin ^{ - 1}}x + 3\sqrt {1 - {x^2}} + c \hfill \\ \hfill \\ \end{gathered}$