If \(\int\displaystyle\frac{(1 + x\log x)}{xe^{-x}}\,dx = e^x f(x) + C\), where C is constant of integration, then \(f(x)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\log x\)
Dividing by \(e^{-x}\) is multiplying by \(e^x\), so the integrand splits as:
\( \displaystyle \frac{1 + x\log x}{x e^{-x}} = e^x\left(\frac{1}{x} + \log x\right) \)
This has the form \(e^x\left(f(x) + f'(x)\right)\) with \(f(x) = \log x\), since \(f'(x) = \frac{1}{x}\). By the standard result \(\int e^x\left(f(x) + f'(x)\right)dx = e^x f(x) + C\):
\( \displaystyle \int e^x\left(\log x + \frac{1}{x}\right)dx = e^x \log x + C \)
So \(f(x) = \log x\).