Answer
\[ = - \frac{1}{x}{\sin ^{ - 1}}\,\left( {ax} \right) - a\ln \left| {1 + \sqrt {1 - {a^2}{x^2}} } \right| + a\ln \left| {ax} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{{\sin }^{ - 1}}ax}}{{{x^2}}}\,\,dx} \hfill \\
\hfill \\
Usi\,ng\,\,table \hfill \\
\hfill \\
\int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}} \,dx \hfill \\
\hfill \\
\int_{}^{} {\frac{1}{{x\sqrt {{a^2} - {x^2}} }}dx = \frac{1}{a}\ln \left| {\frac{{a + \sqrt {{a^2} - {x^2}} }}{x}} \right| + C} \hfill \\
\hfill \\
set\,\,,\,\,\,u = ax\,\,\,\,\,\,then\,\,\,\,\frac{{du}}{a} = dx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{{{\sin }^{ - 1}}ax}}{{{x^2}}}\,\,dx} = a\int_{}^{} {\frac{{{{\sin }^{ - 1}}u}}{{{u^2}}}} du \hfill \\
\hfill \\
= a\,\,\left[ { - \frac{{{{\sin }^{ - 1}}u}}{u} + \int_{}^{} {\frac{1}{{u\sqrt {1 - {u^2}} }}du} } \right] \hfill \\
\hfill \\
= a\,\,\left[ { - \frac{{{{\sin }^{ - 1}}u}}{u} - \ln \left| {\frac{{1 + \sqrt {1 - {u^2}} }}{u}} \right|} \right] + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = ax \hfill \\
\hfill \\
= - \frac{1}{x}{\sin ^{ - 1}}\,\left( {ax} \right) - a\ln \left| {1 + \sqrt {1 - {a^2}{x^2}} } \right| + a\ln \left| {ax} \right| + C \hfill \\
\hfill \\
Also\,\,using\,\,CAS\,\,we\,\,get \hfill \\
\hfill \\
\int_{}^{} {\frac{{{{\sin }^{ - 1}}ax}}{{{x^2}}}\,\,dx} \, = - \frac{1}{x}{\sin ^{ - 1}}\,\left( {ax} \right) - a\ln \left| {1 + \sqrt {1 - {a^2}{x^2}} } \right| + a\ln \left| {ax} \right| + C \hfill \\
\end{gathered} \]
