Answer
$$\eqalign{
& \left( {\text{a}} \right)\frac{1}{2}{\tan ^{ - 1}}\frac{{{e^x}}}{2} + C \cr
& \left( {\text{b}} \right)\frac{1}{2}{\sin ^{ - 1}}\frac{{2x}}{3} + C \cr
& \left( {\text{c}} \right)\frac{{\sqrt 5 }}{5}{\sec ^{ - 1}}\left( {\sqrt {\frac{5}{3}} y} \right) + C \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\int {\frac{{{e^x}}}{{4 + {e^{2x}}}}} dx \cr
& = \int {\frac{{{e^x}}}{{{{\left( 2 \right)}^2} + {{\left( {{e^x}} \right)}^2}}}} dx \cr
& {\text{Apply the formula }}\left( {23} \right){\text{ }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{u}{a} + C \cr
& \int {\frac{{{e^x}}}{{{{\left( 2 \right)}^2} + {{\left( {{e^x}} \right)}^2}}}} dx = \frac{1}{2}{\tan ^{ - 1}}\frac{{{e^x}}}{2} + C \cr
& \cr
& \left( {\text{b}} \right)\int {\frac{{dx}}{{\sqrt {9 - 4{x^2}} }}} \cr
& = \frac{1}{2}\int {\frac{{2dx}}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {2x} \right)}^2}} }}} \cr
& {\text{Apply the formula }}\left( {24} \right){\text{ }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\frac{u}{a} + C \cr
& \frac{1}{2}\int {\frac{{2dx}}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {2x} \right)}^2}} }}} = \frac{1}{2}{\sin ^{ - 1}}\frac{{2x}}{3} + C \cr
& \cr
& \left( {\text{c}} \right)\int {\frac{{dy}}{{y\sqrt {5{y^2} - 3} }}} \cr
& = \int {\frac{{\sqrt 5 dy}}{{\sqrt 5 y\sqrt {{{\left( {\sqrt 5 y} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} }}} \cr
& {\text{Apply the formula }}\left( {25} \right){\text{ }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }}} = \frac{1}{a}{\sec ^{ - 1}}\left| {\frac{u}{a}} \right| + C \cr
& \int {\frac{{\sqrt 5 dy}}{{\sqrt 5 y\sqrt {{{\left( {\sqrt 5 y} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} }}} = \frac{1}{{\sqrt 5 }}{\sec ^{ - 1}}\left| {\frac{{\sqrt 5 y}}{{\sqrt 3 }}} \right| + C \cr
& {\text{ }} = \frac{{\sqrt 5 }}{5}{\sec ^{ - 1}}\left( {\sqrt {\frac{5}{3}} y} \right) + C \cr} $$
