Question

\(\displaystyle \int e^{x}\!\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\,dx=\) ?

• A. \(\dfrac{e^{x}}{x}+k\)
• B. \(-x\,e^{x}+k\)
• C. \(k-\dfrac{e^{x}}{x}\)
• D. \(k-\dfrac{e^{x}}{x^{2}}\)

Hint:

A pattern like $e^x\!\left(\frac{1}{x}-\frac{1}{x^{2}}\right)$ screams "derivative of $\frac{e^x}{x}$".

Answer 1

The Correct Option is A

Differentiate \(\dfrac{e^{x}}{x}\) using quotient rule: \[ \left(\frac{e^{x}}{x}\right)'=\frac{x e^{x}-e^{x}}{x^{2}}=e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right). \] Matches the integrand, so the antiderivative is \(\dfrac{e^{x}}{x}+k\).