Answer
$$\eqalign{
& \left( {\text{a}} \right)\arcsin x + C \cr
& \left( {\text{b}} \right) - \sqrt {1 - {x^2}} + C \cr
& \left( {\text{c}} \right){\text{Cannot be evaluated with basic integration}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{We know that }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = \arcsin \frac{u}{a}} + C,{\text{ then}} \cr
& \int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx = \arcsin x + C,{\text{ with }}u = x \cr
& \cr
& \left( {\text{b}} \right)\int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Integrate by substitution, let }}u = 1 - {x^2},{\text{ }}du = - 2xdx \cr
& \int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx = \int {\frac{1}{{\sqrt u }}} \left( { - \frac{1}{2}} \right)du \cr
& = - \frac{1}{2}\int {{u^{ - 1/2}}du} \cr
& = - {u^{1/2}} + C \cr
& = - \sqrt u + C \cr
& {\text{Write in terms of }}x \cr
& = - \sqrt {1 - {x^2}} + C \cr
& \cr
& \left( {\text{c}} \right)\int {\frac{1}{{x\sqrt {1 - {x^2}} }}} dx \cr
& {\text{This integral cannot be evaluated with basic integration}} \cr
& {\text{rules studied so far}}{\text{.}} \cr} $$
