Use a change of variables to evaluate the following definite integral. \( \int_{2}^{3} \frac{9}{25 x^{2}+40 x+16} d x \)

To evaluate the integral \( \int_{2}^{3} \frac{9}{25 x^{2}+40 x+16} d x \) using a change of variables, we first observe that the denominator can be rewritten in a more manageable form. The expression \( 25 x^{2} + 40 x + 16 \) can be factored or completed as follows: We notice that \( 25 x^{2} + 40 x + 16 \) can be rewritten as: \[ 25(x^{2} + \frac{40}{25}x + \frac{16}{25}) = 25(x^{2} + \frac{8}{5} x + \frac{16}{25}) \] Next, we complete the square: \[ x^{2} + \frac{8}{5}x = (x + \frac{4}{5})^2 - \frac{16}{25} \] Thus: \[ 25\left((x + \frac{4}{5})^2 - \frac{16}{25}\right) + 16 = 25(x + \frac{4}{5})^2 - 16 + 16 = 25(x + \frac{4}{5})^2 \] Now, we have: \[ \int_{2}^{3} \frac{9}{25(x + \frac{4}{5})^2} d x = \frac{9}{25} \int_{2}^{3} \frac{1}{(x + \frac{4}{5})^2} d x \] Now, we perform a change of variables. Let \( u = x + \frac{4}{5} \). Then, \( du = dx \) and our limits change from \( x = 2 \) to \( u = 2 + \frac{4}{5} = \frac{14}{5} \) and \( x = 3 \) to \( u = 3 + \frac{4}{5} = \frac{19}{5} \). Thus: \[ \frac{9}{25} \int_{\frac{14}{5}}^{\frac{19}{5}} \frac{1}{u^2} du \] The integral of \( \frac{1}{u^2} \) is: \[ -\frac{1}{u} + C \] Evaluating: \[ -\frac{9}{25} \left[ \frac{1}{u} \right]_{\frac{14}{5}}^{\frac{19}{5}} = -\frac{9}{25} \left( \frac{5}{19} - \frac{5}{14} \right) \] Finding a common denominator (which would be \( 266 \)): \[ - \frac{9}{25} \left( \frac{70 - 95}{266} \right) = - \frac{9}{25} \left(-\frac{25}{266}\right) = \frac{9}{25} \cdot \frac{25}{266} = \frac{9}{266} \] Thus, the value of the definite integral \( \int_{2}^{3} \frac{9}{25 x^{2}+40 x+16} d x \) is \(\frac{9}{266}\).