\(\displaystyle\int e^x\left(\log x + \frac{1}{x}\right)dx,\ x \gt 0\), is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(e^x\log x + C\)
The integrand has the form \(e^x\left[f(x) + f'(x)\right]\) with
\( \displaystyle f(x) = \log x, \quad f'(x) = \frac{1}{x} \)
Integration by parts gives the standard result
\( \displaystyle \int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C \)
So the integral is \(e^x\log x + C\).
Check by differentiating with the product rule:
\( \displaystyle \frac{d}{dx}\left(e^x\log x\right) = e^x\log x + e^x \times \frac{1}{x} \)
\( \displaystyle = e^x\left(\log x + \frac{1}{x}\right) \)