∫(x+2)/(x^2e^x/2)dx=

The correct answer is Option (4) → \( \displaystyle\frac{-2}{xe^{x/2}}+C \), where \( C \) is an arbitrary constant

The integrand can be written as \( e^{-x/2}\left(\frac{1}{x}+\frac{2}{x^2}\right) \), which is closely related to the derivative of \( \frac{e^{-x/2}}{x} \):

    \( \displaystyle \frac{d}{dx}\left(\frac{e^{-x/2}}{x}\right) \)

    \( \displaystyle =\frac{-\frac{1}{2}e^{-x/2}\cdot x-e^{-x/2}}{x^2} \)

    \( \displaystyle =-\frac{1}{2}e^{-x/2}\left(\frac{1}{x}+\frac{2}{x^2}\right) \)

So the integrand is exactly −2 times this derivative:

    \( \displaystyle \frac{x+2}{x^2e^{x/2}}=e^{-x/2}\cdot\frac{x+2}{x^2} \)

    \( \displaystyle =-2\,\frac{d}{dx}\left(\frac{e^{-x/2}}{x}\right) \)

    \( \displaystyle \int\frac{x+2}{x^2e^{x/2}}\,dx=-2\cdot\frac{e^{-x/2}}{x}+C \)

    \( \displaystyle =\frac{-2}{xe^{x/2}}+C \)

Shortcut: All options have the form \( \frac{k}{xe^{x/2}} \), whose derivative is \( -\frac{k}{2}\cdot\frac{x+2}{x^2e^{x/2}} \). For this to equal the integrand, \( -\frac{k}{2}=1 \), so k = −2.