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
Answer & explanation
Correct answer: option 2
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: $\log x$