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

The fraction can be simplified:

$\frac{x-1}{2x^2} = \frac{x}{2x^2} - \frac{1}{2x^2}$

$= \frac{1}{2x} - \frac{1}{2x^2}$

The integral becomes:

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

The integration formula for $e^x$ is:

$\int e^x[f(x) + f'(x)]dx = e^x \cdot f(x) + C$

When $e^x$ is multiplied by a function plus its derivative, the integral is $e^x$ times that function.

Let $f(x) = \frac{1}{2x}$

The derivative is:

$f'(x) = \frac{d}{dx}\left(\frac{1}{2x}\right)$

$= -\frac{1}{2x^2}$

Check if $f(x) + f'(x)$ matches the expression:

$f(x) + f'(x) = \frac{1}{2x} + \left(-\frac{1}{2x^2}\right)$

$= \frac{1}{2x} - \frac{1}{2x^2}$

This matches the integrand.

Applying the formula with $f(x) = \frac{1}{2x}$:

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

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

Therefore, the integral equals $\frac{e^x}{2x} + C$