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

$\frac{1}{3x}e^x+C$, where C is constant of integration

The correct answer is Option (4) → $\frac{1}{3x}e^x+C$, where C is constant of integration

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

Rewrite:

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

Note that:

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

Therefore:

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

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