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