A) \[{{\tan }^{-1}}\left( \frac{{{e}^{x}}+2}{3} \right)+c\]
B) \[{{\sin }^{-1}}\left( \frac{{{e}^{x}}+2}{3} \right)+c\]
C) \[{{\cos }^{-1}}\left( \frac{{{e}^{x}}+2}{3} \right)+c\]
D) None of the above
Correct Answer: B
Solution :
Let \[I=\int{\frac{{{e}^{x}}\,dx}{\sqrt{5-4{{e}^{x}}-{{e}^{2x}}}}}\] Put\[{{e}^{x}}=t\] \[\Rightarrow \] \[{{e}^{x}}dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{\sqrt{5-4t-{{t}^{2}}}}}\] \[=\int{\frac{dt}{\sqrt{5({{t}^{2}}+4t+4)+4}}}\] \[I=\int{\frac{dt}{\sqrt{9-{{(t+2)}^{2}}}}}\] \[\Rightarrow \] \[I={{\sin }^{-1}}\left( \frac{t+2}{3} \right)+c\] \[\Rightarrow \] \[I={{\sin }^{-1}}\left( \frac{{{e}^{x}}+2}{3} \right)+c\]You need to login to perform this action.
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