Question

$\int e^{x} \frac{\left(x-1\right)}{x^{2}} dx$ is equal to

• A. $\frac{e^{x}}{x^{2}} +c$
• B. $\frac{-e^{x}}{x^{2}} +c$
• C. $\frac{e^{x}}{x} +c$
• D. $\frac{-e^{x}}{x} +c$

Answer 1

Answer: (C)

$\frac{e^{x}}{x} +c$

Answer 2

$\int e^{x}\left(\frac{x-1}{x^{2}}\right) d x$ $=\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x$ $=\frac{e^{x}}{x}+c\left[\because \int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+c\right]$