\( \int e^x\left(\log x+\displaystyle\frac{1}{x^2}\right)dx \) is equal to: ( consider \( \log x=\log_e x \) )
[\( C \) is an arbitrary constant]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( e^x\left(\log x-\displaystyle\frac{1}{x}\right)+c \)
The standard result is \( \int e^x\big(f(x)+f'(x)\big)\,dx=e^x f(x)+c \), so 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 bracket, so
\( \displaystyle \int e^x\left(\log x+\frac{1}{x^2}\right)dx=e^x\left(\log x-\frac{1}{x}\right)+c \)
Why the other options are incorrect
Option (1) \( e^x \log x+c \) — Incorrect. The derivative of \( e^x\log x \) is \( e^x\left(\log x+\frac{1}{x}\right) \), not the given integrand; log x and \( \frac{1}{x^2} \) are not a function and its derivative.
Option (2) \( e^x\displaystyle\frac{1}{x^2}+c \) — Incorrect. The derivative of \( e^x\frac{1}{x^2} \) is \( e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right) \), with no log x term.
Option (4) \( e^x\left(\log x+\displaystyle\frac{1}{x}\right)+c \) — Incorrect. The sign of \( \frac{1}{x} \) is wrong: the derivative of \( e^x\left(\log x+\frac{1}{x}\right) \) is \( e^x\left(\log x+\frac{2}{x}-\frac{1}{x^2}\right) \).