## Calculus: Early Transcendentals (2nd Edition)

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