\( \int _{}^{}x^{-1}\sin (1+\frac{1}{x}) d x \)

To solve the integral \[ I = \int x^{-1} \sin\left(1 + \frac{1}{x}\right) \, dx, \] we can use a substitution to simplify the expression. Let's set \[ u = 1 + \frac{1}{x}. \] Then, we differentiate \(u\) with respect to \(x\): \[ \frac{du}{dx} = -\frac{1}{x^2} \implies du = -\frac{1}{x^2} \, dx \implies dx = -x^2 \, du. \] Next, we need to express \(x\) in terms of \(u\). From our substitution, we have: \[ u - 1 = \frac{1}{x} \implies x = \frac{1}{u - 1}. \] Now, we can substitute \(x\) into the expression for \(dx\): \[ dx = -\left(\frac{1}{u - 1}\right)^2 \, du = -\frac{1}{(u - 1)^2} \, du. \] Now we can substitute \(x^{-1}\) and \(dx\) into the integral: \[ x^{-1} = u - 1, \] so we have: \[ I = \int (u - 1) \sin(u) \left(-\frac{1}{(u - 1)^2}\right) \, du. \] This simplifies to: \[ I = -\int \sin(u) \frac{1}{(u - 1)} \, du. \] Now we can split the integral: \[ I = -\int \frac{\sin(u)}{u - 1} \, du. \] This integral does not have a simple elementary form, but it can be expressed in terms of special functions or evaluated numerically for specific limits. Thus, the integral \[ \int x^{-1} \sin\left(1 + \frac{1}{x}\right) \, dx \] can be expressed as \[ -\int \frac{\sin(u)}{u - 1} \, du, \] which is the final form of the integral.