Answer
$$\frac{{{e^x}}}{2}\sqrt {3 - 4{e^{2x}}} + \frac{3}{4}{\sin ^{ - 1}}\frac{{2{e^x}}}{{\sqrt 3 }} + C$$
Work Step by Step
$$\eqalign{
& \int {{e^x}\sqrt {3 - 4{e^{2x}}} dx} \cr
& {\text{write the integrand as}} \cr
& = \int {{e^x}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {{\left( {2{e^x}} \right)}^2}} dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = 2{e^x},\,\,\,\,\,\,\,du = 2{e^x}dx,\,\,\,\,\,\,dx = \frac{{du}}{{2{e^x}}} \cr
& {\text{write in terms of }}u \cr
& \int {{e^x}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {{\left( {2{e^x}} \right)}^2}} dx} = \int {{e^x}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} \left( {\frac{{du}}{{2{e^x}}}} \right)} \cr
& = \frac{1}{2}\int {\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 74}} \cr
& \left( {74} \right):\,\,\,\,\,\,\int {\sqrt {{a^2} - {u^2}} du = } \frac{u}{2}\sqrt {{a^2} - {u^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{u}{a} + C \cr
& {\text{take }}a = \sqrt 3 ,\,\, \cr
& \frac{1}{2}\int {\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} du} = \frac{u}{4}\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} + \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{4}{\sin ^{ - 1}}\frac{u}{{\sqrt 3 }} + C \cr
& {\text{simplifying}} \cr
& = \frac{u}{2}\sqrt {3 - {u^2}} + \frac{3}{4}{\sin ^{ - 1}}\frac{u}{{\sqrt 3 }} + C \cr
& {\text{write in terms of }}x{\text{; replace }}2{e^x}{\text{ for }}u \cr
& = \frac{{2{e^x}}}{4}\sqrt {3 - 4{e^{2x}}} + \frac{3}{4}{\sin ^{ - 1}}\frac{{2{e^x}}}{{\sqrt 3 }} + C \cr
& = \frac{{{e^x}}}{2}\sqrt {3 - 4{e^{2x}}} + \frac{3}{4}{\sin ^{ - 1}}\frac{{2{e^x}}}{{\sqrt 3 }} + C \cr} $$
