Let \(u = \sqrt{x}\), then \(x = u^2\).

Differentiate to find \(dx\):

\(\frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u}\)

Thus, \(du = \frac{1}{2u} \, dx\) or \(dx = 2u \, du\).

Substitute \(x = u^2\) and \(dx = 2u \, du\) into the integral:

\(I = \int_{1}^{4} \frac{(u) - 1}{2(u^2 + u)} \, 2u \, du\)

\(I = \int_{1}^{4} \frac{u(u - 1)}{2(u^2 + u)} \, 2u \, du\)

\(I = \int_{1}^{4} \frac{u(u - 1)}{u^2 + u} \, du\)

Change the limits of integration:

When \(x = 1\), \(u = \sqrt{1} = 1\).

When \(x = 4\), \(u = \sqrt{4} = 2\).

Thus, \(I = \int_{1}^{2} \frac{u - 1}{u + 1} \, du\).