\(\int e^x\left(\displaystyle\frac{1}{x^2} - \frac{2}{x^3}\right)dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{e^x}{x^2} + C\)
The result to use is
\( \displaystyle \int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C \)
Let \(f(x) = \frac{1}{x^2} = x^{-2}\). Then
\( \displaystyle f'(x) = -2x^{-3} = -\frac{2}{x^3} \)
So the bracket is exactly \(f(x) + f'(x)\), and
\( \displaystyle \int e^x\left(\frac{1}{x^2} - \frac{2}{x^3}\right)dx = \frac{e^x}{x^2} + C \)
Check: \(\frac{d}{dx}\left(\frac{e^x}{x^2}\right) = \frac{e^x}{x^2} - \frac{2e^x}{x^3}\), which is the integrand.