The area under the curve from \(x = 1\) to \(x = 2\) is given by the integral:

\(\int_{1}^{2} \left( 2x + \frac{8}{x^2} \right) \, dx\)

We can split this into two separate integrals:

\(\int_{1}^{2} 2x \, dx + \int_{1}^{2} \frac{8}{x^2} \, dx\)

The first integral is:

\(\int 2x \, dx = x^2\)

Evaluated from 1 to 2:

\([x^2]_{1}^{2} = 2^2 - 1^2 = 4 - 1 = 3\)

The second integral is:

\(\int \frac{8}{x^2} \, dx = -\frac{8}{x}\)

Evaluated from 1 to 2:

\(\left[-\frac{8}{x}\right]_{1}^{2} = -\frac{8}{2} + \frac{8}{1} = -4 + 8 = 4\)

Adding both results gives the total area:

\(3 + 4 = 7\)