If
\(\int(7 x-2) \sqrt{3 x+2} \mathrm{~d} x=\mathrm{A}(3 x+2)^{\frac{5}{2}}+\mathrm{B}(3 x+2)^{\frac{3}{2}}+\mathrm{c}\)
(where c is a constant of integration), then the values of \(A\) and \(B\) are respectively

Let \(\mathrm{I}=\int(7 x-2) \sqrt{3 x+2} \mathrm{~d} x\)
Let \(3 x+2=\mathrm{t} \Rightarrow x=\frac{\mathrm{t}-2}{3} \Rightarrow \mathrm{~d} x=\frac{1}{3} \mathrm{dt}\)
\(\therefore \mathrm{I} \)
\( =\int\left[7\left(\frac{\mathrm{t}-2}{3}\right)-2\right] \sqrt{\mathrm{t}} \mathrm{dt} \)
\( =\frac{1}{3} \int\left(\frac{7 \mathrm{t}}{3}-\frac{20}{3}\right) \sqrt{\mathrm{t}} \mathrm{dt} \)
\( =\frac{7}{9} \int \mathrm{t}^{\frac{3}{2}} \mathrm{dt}-\frac{20}{9} \int \mathrm{t}^{\frac{1}{2}} \mathrm{dt} \)
\( =\frac{14}{45}(3 x+2)^{\frac{5}{2}}-\frac{40}{27}(3 x+2)^{\frac{3}{2}}+\mathrm{C} \)
\( \therefore A=\frac{14}{45} \text { and } B=\frac{-40}{27}\)