The integral ∫ e^x ≤ft((x-1)/(2x^2))dx is equal to

The correct answer is Option (1) → \(\displaystyle\frac{e^x}{2x}+C\), where C is constant of integration.

Splitting the fraction puts the integrand in the form \(e^x\left[f(x) + f'(x)\right]\):

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

With \(f(x) = \displaystyle\frac{1}{2x}\),

    \( \displaystyle f'(x) = -\frac{1}{2x^2} \)

By the standard result \(\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C\),

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