A) \[{{\tan }^{-1}}({{e}^{x}})+c\]
B) \[\log ({{e}^{x}}+{{e}^{-x}})+c\]
C) \[\log ({{e}^{2x}}+1)+c\]
D) None of these
Correct Answer: A
Solution :
Let\[I=\int{\frac{dx}{{{e}^{x}}+{{e}^{-x}}}}=\int{\frac{{{e}^{x}}}{{{e}^{2x}}+1}}dx\] Let \[{{e}^{x}}=t\] \[\Rightarrow \] \[{{e}^{x}}dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{{{t}^{2}}+1}}\] \[\Rightarrow \] \[I={{\tan }^{-1}}t+c\] \[\Rightarrow \] \[I={{\tan }^{-1}}{{e}^{x}}+c\]
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