\( \int \frac { x ^ { 2 } - 4 } { x - 2 } \partial x = \)

To solve the integral \[ \int \frac{x^2 - 4}{x - 2} \, dx, \] we first simplify the integrand. Notice that the expression \(x^2 - 4\) can be factored: \[ x^2 - 4 = (x - 2)(x + 2). \] Now we can rewrite the integrand: \[ \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}. \] For \(x \neq 2\), we can cancel \(x - 2\): \[ \frac{x^2 - 4}{x - 2} = x + 2. \] Thus, the integral simplifies to: \[ \int (x + 2) \, dx. \] Now we can integrate term by term: \[ \int (x + 2) \, dx = \int x \, dx + \int 2 \, dx. \] Calculating these integrals gives: \[ \int x \, dx = \frac{x^2}{2} + C_1, \] \[ \int 2 \, dx = 2x + C_2. \] Combining these results, we have: \[ \int (x + 2) \, dx = \frac{x^2}{2} + 2x + C, \] where \(C = C_1 + C_2\) is the constant of integration. Thus, the final result is: \[ \int \frac{x^2 - 4}{x - 2} \, dx = \frac{x^2}{2} + 2x + C. \]