The value of \[\int_{{}}^{{}}{\left( 1+\frac{1}{{{x}^{2}}} \right)\ {{e}^{\left( x-\frac{1}{x} \right)}}}\ dx\] equals [Kurukshetra CEE 1998]

\[{{e}^{{{x}^{2}}-\frac{1}{x}}}+c\]

\[{{e}^{{{x}^{2}}+\frac{1}{{{x}^{2}}}}}+c\]

The value of \[\int_{{}}^{{}}{\left( 1+\frac{1}{{{x}^{2}}} \right)\ {{

1.	The value of \[\int_{{}}^{{}}{\left( 1+\frac{1}{{{x}^{2}}} \right)\ {{e}^{\left( x-\frac{1}{x} \right)}}}\ dx\] equals [Kurukshetra CEE 1998]
A.	\[{{e}^{x-\frac{1}{x}}}+c\]
B.	\[{{e}^{x+\frac{1}{x}}}+c\]
C.	\[{{e}^{{{x}^{2}}-\frac{1}{x}}}+c\]
D.	\[{{e}^{{{x}^{2}}+\frac{1}{{{x}^{2}}}}}+c\]
Answer» B. \[{{e}^{x+\frac{1}{x}}}+c\]

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The value of \[\int_{{}}^{{}}{\left( 1+\frac{1}{{{x}^{2}}} \right)\ {{e}^{\left( x-\frac{1}{x} \right)}}}\ dx\] equals [Kurukshetra CEE 1998]

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