\( \int e^x\left(\log x+\displaystyle\frac{1}{x^2}\right)dx \) is equal to
(consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( e^x\left(\log_e x-\displaystyle\frac{1}{x}\right)+c \) : where c is an arbitrary constant
By the standard result
\( \displaystyle \int e^x\left[f(x)+f'(x)\right]dx=e^x f(x)+c \)
the bracket has to be written as a function plus its derivative. Let \( f(x)=\log x-\frac{1}{x} \). Then
\( \displaystyle f'(x)=\frac{1}{x}+\frac{1}{x^2} \)
\( \displaystyle f(x)+f'(x)=\log x-\frac{1}{x}+\frac{1}{x}+\frac{1}{x^2} \)
\( \displaystyle =\log x+\frac{1}{x^2} \)
This is exactly the given bracket, so
\( \displaystyle \int e^x\left(\log x+\frac{1}{x^2}\right)dx=e^x\left(\log_e x-\frac{1}{x}\right)+c \)
Why the other options are incorrect
Option (1) \( e^x\log_e x+c \) : where \( c \) is an arbitrary constant — Incorrect. The derivative of \( e^x\log x \) is \( e^x\left(\log x+\frac{1}{x}\right) \), not \( e^x\left(\log x+\frac{1}{x^2}\right) \).
Option (2) \( e^x\left(\log_e x+\displaystyle\frac{1}{x}\right)+c \) : where c is an arbitrary constant — Incorrect. The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \); differentiating this option gives \( e^x\left(\log x+\frac{2}{x}-\frac{1}{x^2}\right) \).
Option (4) \( e^x\cdot\displaystyle\frac{1}{x}+c \) : where \( c \) is an arbitrary constant — Incorrect. The derivative of \( e^x\cdot\frac{1}{x} \) is \( e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) \), which has no \( \log x \) term.