Answer:
Let \[l=\int{\frac{{{e}^{x}}}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}\,dx}\] Now, put \[{{e}^{x}}=t\] \[\Rightarrow \] \[{{e}^{x}}dx=dt\] \[\therefore \] \[l=\int{\frac{dt}{\sqrt{5-4t-{{t}^{2}}}}}\] \[=\int{\frac{dt}{\sqrt{-({{t}^{2}}+4t-5)}}}\] \[=\int{\frac{dt}{\sqrt{-[{{(t\,+2)}^{2}}-5-4]}}}\] [using completing the square] \[=\int{\frac{dt}{\sqrt{9-{{(t\,+2)}^{2}}}}}\] \[=\int{\frac{dt}{\sqrt{{{(3)}^{2}}-{{(t\,+2)}^{2}}}}}\] \[={{\sin }^{-1}}\left( \frac{t+2}{3} \right)+C\] \[\left[ \because \,\,\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\frac{x}{a}} \right]\] \[={{\sin }^{-1}}\left( \frac{{{e}^{x}}+2}{3} \right)+C\] \[[\because \,\,t={{e}^{x}}]\]
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