If int(7x-2)sqrt{3x+2} , dx = A(3x+2)^{frac{5 | vedclass

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(D) Let $I = \int(7x-2)\sqrt{3x+2} \, dx$. Substitute $3x+2 = t$,so $x = \frac{t-2}{3}$ and $dx = \frac{1}{3} dt$. Substituting these into the integra

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$\frac{14}{45}, \frac{-40}{27}$

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(D) Let $I = \int(7x-2)\sqrt{3x+2} \, dx$. Substitute $3x+2 = t$,so $x = \frac{t-2}{3}$ and $dx = \frac{1}{3} dt$. Substituting these into the integral: $I = \int \left[7\left(\frac{t-2}{3}\right) - 2\right] \sqrt{t} \cdot \frac{1}{3} dt$ $I = \frac{1}{3} \int \left(\frac{7t - 14 - 6}{3}\right) \sqrt{t} \, dt$ $I = \frac{1}{9} \int (7t - 20) t^{\frac{1}{2}} \, dt$ $I = \frac{7}{9} \int t^{\frac{3}{2}} \, dt - \frac{20}{9} \int t^{\frac{1}{2}} \, dt$ $I = \frac{7}{9} \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}} - \frac{20}{9} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c$ $I = \frac{14}{45} t^{\frac{5}{2}} - \frac{40}{27} t^{\frac{3}{2}} + c$ Substituting $t = 3x+2$ back: $I = \frac{14}{45}(3x+2)^{\frac{5}{2}} - \frac{40}{27}(3x+2)^{\frac{3}{2}} + c$ Comparing this with the given form,we get $A = \frac{14}{45}$ and $B = \frac{-40}{27}$.

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

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