Question

Integrate the function: $e^x(\frac{1}{x}-\frac{1}{x^2})$

Answer 1

The correct answer is: $I=\frac{e^x}{x}+C$
Let $I=∫e^x[\frac{1}{x}-\frac{1}{x^2}]dx$
Also,let $\frac{1}{x}=ƒ(x)\,ƒ'(x)=\frac{-1}{x^2}$
It is known that,$∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C$
$∴I=\frac{e^x}{x}+C$