\(\int e^x\left(\log x+\displaystyle\frac{1}{x}\right)dx=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(e^x\log x+C\)
The integrand has the form \(e^x\left[f(x) + f'(x)\right]\), and
\( \int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C \)
Here
\( \displaystyle f(x) = \log x,\quad f'(x) = \frac{1}{x} \)
So
\( \displaystyle \int e^x\left(\log x + \frac{1}{x}\right)dx = e^x\log x + C \)