Q.Choose the correct option: $\int \frac{e^x(x-1)}{x^2}dx =$
(A) $\frac{e^x}{x}+c$ (B) $\frac{e^x}{x^2}+c$ (C) $\left(x-\frac1x\right)e^x+c$ (D) $xe^{-x}+c$

Because $\frac{d}{dx}[e^xf(x)]=e^xf(x)+e^xf'(x)=e^x[f(x)+f'(x)]$, any integrand that can be written as $e^x$ multiplied by "some function $f(x)$ plus its own derivative $f'(x)$" integrates immediately to $e^xf(x)+c$, without any explicit by-parts computation. The entire skill lies in recognising — or algebraically manufacturing, by splitting a fraction, factoring a numerator, or substituting an inner variable — the exact pair $f(x)$ and $f'(x)$ hiding inside a more complicated-looking expression multiplying $e^x$ (or, after a subst …