A) \[\frac{-{{e}^{-x}}}{({{e}^{x}}+{{e}^{-x}})}+c\]
B) \[\frac{-1}{({{e}^{x}}+{{e}^{-x}})}+c\]
C) \[\frac{1}{{{({{e}^{x}}+1)}^{2}}}+c\]
D) \[\frac{1}{{{({{e}^{x}}+{{e}^{-x}})}^{2}}}+c\]
E) \[\frac{1}{({{e}^{x}}-{{e}^{-x}})}+c\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{2dx}{{{({{e}^{x}}+{{e}^{-x}})}^{2}}}}\] \[=\int{\frac{2dx}{{{e}^{2x}}+{{e}^{-2x}}+2}}\] \[=\int{\frac{2{{e}^{2x}}dx}{{{e}^{4x}}+2{{e}^{2x}}+1}}\] Put \[{{e}^{2x=t}}\] \[\Rightarrow \] \[2{{e}^{2x}}dx=dt\] \[\therefore \]\[I=\int{\frac{dt}{{{t}^{2}}+2t+1}}=\int{\frac{dt}{{{(t+1)}^{2}}}}=-\frac{1}{t+1}+c\] \[=-\frac{1}{{{e}^{2x}}+1}=-\frac{{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+c\]
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