A) \[{{\tan }^{-1}}({{e}^{-x}})\]
B) \[{{\tan }^{-1}}({{e}^{x}})\]
C) \[\log ({{e}^{x}}-{{e}^{-x}})\]
D) \[\log ({{e}^{x}}+{{e}^{-x}})\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dx}{{{e}^{x}}+{{e}^{-x}}}}=\int_{{}}^{{}}{\frac{{{e}^{x}}}{{{e}^{2x}}+1}}\,dx=\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+1}}={{\tan }^{-1}}(t)\] \[={{\tan }^{-1}}({{e}^{x}})+c\], {Putting \[{{e}^{x}}=t\Rightarrow {{e}^{x}}dx=dt\}.\]
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