Answer:
Option B
Explanation:
$\int_{}^{} (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$
= $\int_{}^{} e^{x+\frac{1}{x}}dx+\int_{}^{} x\left(1-\frac{1}{x^{2}}\right)e^{x+\frac{1}{x}}dx$
= $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} \frac{d}{dx}(x)e^{x+1/x} dx$
= $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} e^{x+\frac{1}{x}}dx$
$\left[\because \int_{}^{} \left(1-\frac{1}{x^{2}}\right)e^{x+1/x}dx=e^{x+1/x}\right]$
$=\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} ex^{+1/x}dx$
= $xe^{x+1/x}+C$
