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