A) \[-\frac{1}{2({{e}^{2x}}+1)}+c\]
B) \[\frac{1}{2({{e}^{2x}}+1)}+c\]
C) \[-\frac{1}{{{e}^{2x}}+1}\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{1}{{{({{e}^{x}}+{{e}^{-x}})}^{2}}}\,dx=\int_{{}}^{{}}{\frac{{{e}^{2x}}}{{{({{e}^{2x}}+1)}^{2}}}}\,dx}\] Put \[{{e}^{2x}}+1=t\Rightarrow 2{{e}^{2x}}dx=dt,\] then it reduces to \[\frac{1}{2}\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}dt}=-\frac{1}{2}.\frac{1}{t}+c=-\frac{1}{2({{e}^{2x}}+1)}+c.\]
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