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

$\frac{e^x}{2x}+C$, where C is constant of integration.

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

Let the integral be

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

Rewrite the integrand:

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

Now check derivative of $\displaystyle \frac{e^x}{2x}$:

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

This matches the integrand.

Final answer:

$\displaystyle \frac{e^x}{2x} + C$