Answer
$$\left( {\bf{a}} \right){\sin ^{ - 1}}\frac{x}{3} + C,\,\,\,\left( {\bf{b}} \right)\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 5 }} + C,\,\,\,\left( {\bf{c}} \right)\frac{1}{{\sqrt \pi }}{\sec ^{ - 1}}\left| {\frac{x}{{\sqrt \pi }}} \right| + C$$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)\int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} \cr
& {\text{Integrate using the formula }}\left( {24} \right)\,\,\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\frac{u}{a} + C} {\text{. with }}u = x{\text{ and }}a = 3 \cr
& \int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} = {\sin ^{ - 1}}\frac{x}{3} + C \cr
& \cr
& \left( {\bf{b}} \right)\int {\frac{{dx}}{{5 + {x^2}}}} \cr
& \int {\frac{{dx}}{{5 + {x^2}}}} = \int {\frac{{dx}}{{{{\left( {\sqrt 5 } \right)}^2} + {x^2}}}} \cr
& {\text{Integrate using the formula }}\left( {23} \right)\,\,\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{u}{a} + C} {\text{. with }}u = x{\text{ and }}a = \sqrt 5 \cr
& \int {\frac{{dx}}{{{{\left( {\sqrt 5 } \right)}^2} + {x^2}}}} = \frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 5 }} + C \cr
& \cr
& \left( {\bf{c}} \right)\int {\frac{{dx}}{{x\sqrt {{x^2} - \pi } }}} \cr
& \int {\frac{{dx}}{{x\sqrt {{x^2} - \pi } }}} = \int {\frac{{dx}}{{x\sqrt {{x^2} - \left( {\sqrt \pi } \right)} }}} \cr
& {\text{Integrate using the formula }}\left( {25} \right)\,\,\int {\frac{{dx}}{{x\sqrt {{x^2} - \left( {\sqrt \pi } \right)} }} = \frac{1}{a}{{\sec }^{ - 1}}\left| {\frac{u}{a}} \right| + C} {\text{. With }}u = x{\text{ and }}a = \sqrt \pi \cr
& \int {\frac{{dx}}{{x\sqrt {{x^2} - \left( {\sqrt \pi } \right)} }}} = \frac{1}{{\sqrt \pi }}{\sec ^{ - 1}}\left| {\frac{x}{{\sqrt \pi }}} \right| + C \cr} $$
