Answer
\[ = {\sin ^{ - 1}}\,\left( {\frac{{x + 1}}{2}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\sqrt {3 - 2x - {x^2}} }}} \hfill \\
\hfill \\
completing\,the\,square \hfill \\
\hfill \\
3 - 2x - {x^2} = - \,\left( {{x^2} + 2x - 3} \right) \hfill \\
= - \,\left( {{x^2} + 2x + 1 - 4} \right) \hfill \\
= 4 - \,\left( {{x^2} + 2x + 1} \right) \hfill \\
= 4 - \,{\left( {x + 1} \right)^2} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {3 - 2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {4 - \,{{\left( {x + 1} \right)}^2}} }}} \hfill \\
\hfill \\
the\,\,\,integral\,contains\,\,the\,\,form\,\,\,{a^2} - {u^2}\, \hfill \\
\hfill \\
set{\text{ }}\sin \theta = \frac{{x + 1}}{2}{\text{ and }}x = 2\sin \theta - 1 \hfill \\
\hfill \\
then \hfill \\
\hfill \\
dx = 2\cos \theta d\theta \hfill \\
\hfill \\
\sqrt {3 - 2x - {x^2}} = 2\cos \theta \hfill \\
\hfill \\
Substituting{\text{ }}these{\text{ }}factors{\text{ }} \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {4 - \,{{\left( {x + 1} \right)}^2}} }}} = \int_{}^{} {\frac{{2\cos \theta d\theta }}{{2\,\cos \theta }}} = \int_{}^{} {d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \theta + C \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= {\sin ^{ - 1}}\,\left( {\frac{{x + 1}}{2}} \right) + C \hfill \\
\end{gathered} \]
