Lời giải:Ta có: \(\left( {5 + 4x - {x^2}} \right)' = 4 - 2x\) và \(4x - 3 = 5 - 2\left( {4 - 2x} \right)\).
\(I = \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{{4x - 3}}{{\sqrt {5 + 4x - {x^2}} }}dx} \\= \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{5}{{\sqrt {5 + 4x - {x^2}} }}dx} - \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{{2\left( {4 - 2x} \right)}}{{\sqrt {5 + 4x - {x^2}} }}dx} \)
Xét \({I_1} = \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{5}{{\sqrt {5 + 4x - {x^2}} }}dx} = \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{5}{{\sqrt {9 - {{\left( {x - 2} \right)}^2}} }}dx} \).
Đặt \(x - 2 = 3\sin t,t \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow dx = 3\cos tdt\).
Đổi cận \(\left\{ \begin{array}{l}
x = \frac{7}{2} \Rightarrow t = \frac{\pi }{6}\\
x = \frac{1}{2} \Rightarrow t = - \frac{\pi }{6}
\end{array} \right.\).
\( \Rightarrow {I_1} = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{5.3\cos t}}{{\sqrt {9 - 9{{\sin }^2}t} }}dt} = \frac{{5\pi }}{3}\).
Xét \({I_2} = \int\limits_{\frac{1}{2}}^{\frac{7}{2}} {\frac{{2\left( {4 - 2x} \right)}}{{\sqrt {5 + 4x - {x^2}} }}dx} \).
Đặt \(t = 5 + 4x - {x^2} \Rightarrow dt = 4 - 2x\).
Đổi cận \(\left\{ \begin{array}{l}
x = \frac{1}{2} \Rightarrow t = \frac{{27}}{4}\\
x = \frac{7}{2} \Rightarrow t = \frac{{27}}{4}
\end{array} \right. \Rightarrow {I_2} = 0\).
\(\Rightarrow I = \frac{{5\pi }}{3}\).
