Answer
\[ = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} \hfill \\
\hfill \\
integration\,\,by\,\,parts\,\,formula \hfill \\
\hfill \\
\int_{}^{} {udv = uv - \int_{}^{} {vdu} } \hfill \\
\hfill \\
set \hfill \\
u = {\sin ^{ - 1}}x\,\,\,\,then\,\,\,du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \hfill \\
\hfill \\
dv = {x^n}dx\,\,\,\,\,\,then\,\,\,\,\,\,\,v = \frac{{{x^{n + 1}}}}{{n + 1}} \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \int_{}^{} {\frac{{{x^{n + 1}}}}{{n + 1}}} \,\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx \hfill \\
\hfill \\
= \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\
\end{gathered} \]
