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