1.

\[\int_{{}}^{{}}{\frac{dx}{{{e}^{-2x}}{{({{e}^{2x}}+1)}^{2}}}=}\]

A.            \[\frac{-1}{2({{e}^{2x}}+1)}+c\]
B.            \[\frac{1}{2({{e}^{2x}}+1)}+c\]
C.            \[\frac{1}{{{e}^{2x}}+1}+c\]
D.            \[\frac{-1}{{{e}^{2x}}+1}+c\]
Answer» B.            \[\frac{1}{2({{e}^{2x}}+1)}+c\]


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