The integral I = $\int e^x\left(\frac{x - 1}{3x^2}\right) dx$ is equal to

The integral $I = \int e^x\left(\frac{x - 1}{3x^2}\right) dx$

Simplify the fraction:

$\frac{x - 1}{3x^2}$

$= \frac{1}{3} \cdot \frac{x - 1}{x^2}$

$= \frac{1}{3}\left(\frac{x}{x^2} - \frac{1}{x^2}\right)$

$= \frac{1}{3}\left(\frac{1}{x} - \frac{1}{x^2}\right)$

The integral becomes:

$I = \frac{1}{3}\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right) dx$

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For integrals of the form $\int e^x[f(x) + f'(x)]dx = e^x \cdot f(x) + C$

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

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

Therefore:

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

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

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Applying the formula:

$I = \frac{1}{3}\int e^x\left[\frac{1}{x} - \frac{1}{x^2}\right] dx$

$= \frac{1}{3} \cdot e^x \cdot \frac{1}{x} + C$

$= \frac{e^x}{3x} + C$

Therefore, $I = \frac{e^x}{3x} + C$ where $C$ is the constant of integration.