The primitive of the function
$f(x)=\left(1-\frac{1}{x^2}\right) a^{x+\frac{1}{x}}, x>0$, is
Answer & explanation
Correct answer: option 1
The primitive of f(x) is
$\int\left(1-\frac{1}{x^2}\right) a^{x+\frac{1}{x}} d x=\int a^{x+\frac{1}{x}} d\left(x+\frac{1}{x}\right)=\frac{a^{x+\frac{1}{x}}}{\log _e a}+C$