The primitive of the function  $f(x)=\left(1-\frac{1}{x^2}\right) a^{x+\frac{1}{x}}, x>0$, is

$\frac{a^{x+\frac{1}{x}}}{\log _e a}$ $a^{x+\frac{1}{x}} \log _e a$ $\frac{a^{x+\frac{1}{x}}}{x} \log _e a$ $\frac{x a^{x+\frac{1}{x}}}{\log _e a}$

$\frac{a^{x+\frac{1}{x}}}{\log _e a}$

The primitive of f(x) is $\int\left(1-\frac{1}{x^2}\right) a^{x+\frac{1}{x}} d x=\int a^{x+\frac{1}{x}} d\left(x+\frac{1}{x}\right)=\frac{a^{x+\frac{1}{x}}}{\log _e a}+C$