A) \[\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+c\]
B) \[\log |{{e}^{x}}+\sqrt{{{e}^{2x}}-1}|+c\]
C) \[-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}|+c\]
D) \[-\log |{{e}^{-2x}}+\sqrt{{{e}^{-2x}}-1}|+c\]
E) \[\log |{{e}^{-2x}}+\sqrt{{{e}^{-2x}}-1}|+c\]
Correct Answer: C
Solution :
Let \[I=\int{\frac{dx}{\sqrt{1-{{e}^{2x}}}}}=\int{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}}dx\] Put \[t={{e}^{-x}}\] \[\Rightarrow \] \[dt=-{{e}^{-x}}dx\] \[\therefore \]\[I=\int{-\frac{dt}{\sqrt{{{t}^{2}}-1}}}=-\log |t+\sqrt{{{t}^{2}}+1}|+c\] \[=-\log |{{e}^{-x}}+\sqrt{{{e}^{-2x}}-1}+c\]
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