A) \[\frac{-1}{{{e}^{x}}+{{e}^{-x}}}+c\]
B) \[\frac{-{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}}+c\]
C) \[\frac{1}{{{e}^{x}}+{{e}^{-x}}}+c\]
D) \[\frac{{{e}^{x}}}{{{e}^{x}}-{{e}^{-x}}}\]
Correct Answer: B
Solution :
[b] \[I=\int{\frac{2}{{{({{e}^{x}}+{{e}^{-x}})}^{2}}}.dx}=\int{\frac{2.{{e}^{2x}}}{{{({{e}^{2x}}+1)}^{2}}}.dx}\] Let \[z=1+{{e}^{2x}}~~~~~~\Rightarrow dz=2.{{e}^{2x}}.dx\] Now, \[I=\int{\frac{dz}{{{z}^{2}}}}=\int{{{z}^{-2}}.dz}=\frac{{{z}^{-2+1}}}{-2+1}+c=\frac{-1}{{{e}^{2x}}+1}+c=\frac{-{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}}+c\] Hence, option [b] is correct.
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