$\int e^x \left( \dfrac{1}{x} - \dfrac{1}{x^2} \right)$ $dx$

The correct answer is $ \dfrac{e^x}{x}$ + $C$

We know that $\int e^x [f(x) + f ’(x)]$ $dx$ = $e^x f(x)$ + $C$


The given integrand is of the form $e^x [f(x) + f ’(x)]$
where $f(x)$ = $\dfrac{1}{x}$ and $f'(x)$ = $-\dfrac{1}{x^2}$


∴ $\int e^x \left( \dfrac{1}{x} - \dfrac{1}{x^2} \right)$ $dx$ = $e^x \dfrac{1}{x}$ + $C$ = $ \dfrac{e^x}{x}$ + $C$