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

Differentiating both sides, \(\frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u}\), so \(dx = 2u \, du\).

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

\(I = \int_1^4 \frac{1}{u^2(4-u)} \, (2u \, du)\).

This simplifies to:

\(I = \int_1^2 \frac{2}{u(4-u)} \, du\).