79. \( \int_{1}^{9} \frac{1}{\sqrt{x}(1+\sqrt{x})^{2}} d x \)

To solve the integral \( I = \int_{1}^{9} \frac{1}{\sqrt{x}(1+\sqrt{x})^{2}} d x \), we can use a substitution to simplify the expression. Let's set \( u = \sqrt{x} \). Then, \( x = u^2 \) and \( dx = 2u \, du \). The limits of integration change as follows: - When \( x = 1 \), \( u = \sqrt{1} = 1 \). - When \( x = 9 \), \( u = \sqrt{9} = 3 \). Substituting these into the integral gives: \[ I = \int_{1}^{3} \frac{1}{u(1+u)^{2}} (2u \, du) = 2 \int_{1}^{3} \frac{1}{(1+u)^{2}} \, du \] Now, we simplify the integral: \[ I = 2 \int_{1}^{3} (1+u)^{-2} \, du \] The integral of \( (1+u)^{-2} \) can be computed as follows: \[ \int (1+u)^{-2} \, du = -\frac{1}{1+u} + C \] Thus, we can evaluate the definite integral: \[ I = 2 \left[ -\frac{1}{1+u} \right]_{1}^{3} \] Calculating the limits: \[ = 2 \left( -\frac{1}{1+3} + \frac{1}{1+1} \right) = 2 \left( -\frac{1}{4} + \frac{1}{2} \right) \] \[ = 2 \left( -\frac{1}{4} + \frac{2}{4} \right) = 2 \left( \frac{1}{4} \right) = \frac{1}{2} \] Therefore, the value of the integral is: \[ \int_{1}^{9} \frac{1}{\sqrt{x}(1+\sqrt{x})^{2}} d x = \frac{1}{2} \]