Answer
$$2\sqrt {{e^t} - 3} - 2\sqrt 3 {\tan ^{ - 1}}\sqrt {\frac{{{e^t} - 3}}{3}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {{e^t} - 3} } dt \cr
& {\text{Let }}u = \sqrt {{e^t} - 3} ,{\text{ }}{u^2} = {e^t} - 3,{\text{ }}2udu = {e^t}dt \cr
& {\text{Substituting}} \cr
& \int {\sqrt {{e^t} - 3} } dt = \int {u\left( {\frac{{2u}}{{{e^t}}}} \right)} dt \cr
& = \int {u\left( {\frac{{2u}}{{{u^2} + 3}}} \right)} dt \cr
& = \int {\frac{{2{u^2}}}{{{u^2} + 3}}} dt \cr
& {\text{By long division }}\frac{{2{u^2}}}{{{u^2} + 3}} = 2 - \frac{6}{{{u^2} + 3}} \cr
& = \int {\left( {2 - \frac{6}{{{u^2} + 3}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2u - 6\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right)} \right) + C \cr
& = 2u - \frac{6}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}\sqrt {{e^t} - 3} {\text{ for }}u \cr
& = 2\sqrt {{e^t} - 3} - \frac{6}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt {{e^t} - 3} }}{{\sqrt 3 }}} \right) + C \cr
& = 2\sqrt {{e^t} - 3} - 2\sqrt 3 {\tan ^{ - 1}}\sqrt {\frac{{{e^t} - 3}}{3}} + C \cr} $$
