\( \int e^{x}\left(\displaystyle\frac{x-1}{2x^{2}}\right)dx \) is equal to ( where \( C \) is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{e^{x}}{2x}+C \)
Splitting the fraction,
\( \displaystyle \frac{x-1}{2x^{2}}=\frac{1}{2x}-\frac{1}{2x^{2}} \)
With \( f(x)=\frac{1}{2x} \), we have \( f'(x)=-\frac{1}{2x^{2}} \), so the integrand is \( e^{x}\left[f(x)+f'(x)\right] \).
Since \( \int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+C \),
\( \displaystyle \int e^{x}\left(\frac{x-1}{2x^{2}}\right)dx=\frac{e^{x}}{2x}+C \)
Shortcut: Differentiating Option (3) gives back the integrand:
\( \displaystyle \frac{d}{dx}\left(\frac{e^{x}}{2x}\right)=\frac{e^{x}}{2x}-\frac{e^{x}}{2x^{2}} \)
\( \displaystyle =\frac{e^{x}(x-1)}{2x^{2}} \)