Question

$\int e^x\frac{(x-1)}{x^2}dx$ 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

The correct option is C $\frac{e^x}{x}+c$

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