int(2 x-3) sqrt{3 x+2} , dx = | vedclass

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(C) Let $I = \int(2 x-3) \sqrt{3 x+2} \, dx$. Substitute $3x+2 = t^2$, which implies $x = \frac{t^2-2}{3}$ and $dx = \frac{2t}{3} \, dt$. Substituting

Vedclass · Accepted Answer

$\frac{2}{135}(54 x^2-123 x-106) \sqrt{3 x+2}+c$

Vedclass · Answer

(C) Let $I = \int(2 x-3) \sqrt{3 x+2} \, dx$. Substitute $3x+2 = t^2$, which implies $x = \frac{t^2-2}{3}$ and $dx = \frac{2t}{3} \, dt$. Substituting these into the integral: $I = \int \left(2 \left(\frac{t^2-2}{3}\right) - 3\right) \cdot t \cdot \frac{2t}{3} \, dt$ $I = \int \left(\frac{2t^2-4-9}{3}\right) \cdot \frac{2t^2}{3} \, dt = \frac{2}{9} \int (2t^2-13)t^2 \, dt$ $I = \frac{2}{9} \int (2t^4 - 13t^2) \, dt = \frac{2}{9} \left( \frac{2t^5}{5} - \frac{13t^3}{3} \right) + c$ $I = \frac{2}{135} (6t^5 - 65t^3) + c = \frac{2}{135} t^3 (6t^2 - 65) + c$ Since $t = \sqrt{3x+2}$, we have $t^3 = (3x+2)\sqrt{3x+2}$ and $t^2 = 3x+2$. $I = \frac{2}{135} (3x+2) \sqrt{3x+2} (6(3x+2) - 65) + c$ $I = \frac{2}{135} \sqrt{3x+2} (3x+2) (18x + 12 - 65) + c$ $I = \frac{2}{135} \sqrt{3x+2} (3x+2) (18x - 53) + c$ $I = \frac{2}{135} \sqrt{3x+2} (54x^2 - 159x + 36x - 106) + c$ $I = \frac{2}{135} (54x^2 - 123x - 106) \sqrt{3x+2} + c$.

$\int(2 x-3) \sqrt{3 x+2} \, dx =$

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