If $\int\frac{(1 + x\log x)}{xe^{-x}}dx = e^x f(x) + C$, where C is constant of integration, then f(x) is

$\log x$

The correct answer is Option (2) → $\log x$

Given:

$\displaystyle \int \frac{1+x\log x}{x e^{-x}}\,dx = e^{x} f(x) + C$

Simplify integrand:

$\frac{1+x\log x}{x e^{-x}} = e^{x}\left(\frac{1}{x}+\log x\right)$

Thus:

$\displaystyle \int e^{x}\left(\frac{1}{x}+\log x\right)dx = e^{x}f(x)$

Use reverse product rule: If $\frac{d}{dx}(e^{x}f(x)) = e^{x}f(x) + e^{x}f'(x)$, then

$e^{x}\left(\frac{1}{x}+\log x\right) = e^{x}(f(x)+f'(x))$

Divide by $e^{x}$:

$\frac{1}{x}+\log x = f(x)+f'(x)$

Try $f(x)=\log x$:

$f'(x)=\frac{1}{x}$

Then

$f(x)+f'(x)=\log x+\frac{1}{x}$

Matches the RHS, so $f(x)$ is $\log x$.

Final answer: $\log x$