Dividing both the numerator and denominator by $x^2$ gives

$$I=\int \frac{\frac{\cos{x}}{x}-\frac{\sin{x}}{x^2}}{1+\left(\frac{\sin{x}}{x}\right)^2} dx$$

Let $y=\frac{\sin{x}}{x}$, then

$$\frac{dy}{dx}=\frac{-\sin{x} + x\cos{x}}{x^2}=\frac{\cos{x}}{x}-\frac{\sin{x}}{x^2}$$

Then

$$I=\int \frac{dy}{1+y^2}=\arctan{y}+C=\boxed{\arctan{\left(\frac{\sin{x}}{x}\right)}+C}$$