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