A) \[{{e}^{x-\frac{1}{x}}}+c\]
B) \[{{e}^{x+\frac{1}{x}}}+c\]
C) \[{{e}^{{{x}^{2}}-\frac{1}{x}}}+c\]
D) \[{{e}^{{{x}^{2}}+\frac{1}{{{x}^{2}}}}}+c\]
Correct Answer: A
Solution :
\[I=\int_{{}}^{{}}{\left( 1+\frac{1}{{{x}^{2}}} \right)\text{ }{{e}^{x-\frac{1}{x}}}dx}\]. Put \[x-\frac{1}{x}=t\Rightarrow \left( 1+\frac{1}{{{x}^{2}}} \right)\,dx=dt\] \[\therefore \,\,\,I=\int_{{}}^{{}}{{{e}^{t}}dt={{e}^{t}}+c={{e}^{x-\frac{1}{x}}}+c}\].
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