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