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

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(B) Let $I = \int(x+2) \sqrt{x+3} \, dx$. Substitute $t = \sqrt{x+3}$, which implies $t^2 = x+3$, so $x = t^2-3$. Differentiating both sides, $dx = 2t

Vedclass · Accepted Answer

$\frac{2}{15} \sqrt{x+3}(3x^2+13x+12)+C$

Vedclass · Answer

(B) Let $I = \int(x+2) \sqrt{x+3} \, dx$. Substitute $t = \sqrt{x+3}$, which implies $t^2 = x+3$, so $x = t^2-3$. Differentiating both sides, $dx = 2t \, dt$. Substituting these into the integral: $I = \int(t^2-3+2) \cdot t \cdot (2t \, dt)$ $I = \int(t^2-1) \cdot 2t^2 \, dt$ $I = 2 \int(t^4-t^2) \, dt$ $I = 2 \left( \frac{t^5}{5} - \frac{t^3}{3} \right) + C$ $I = 2 \left( \frac{3t^5 - 5t^3}{15} \right) + C$ $I = \frac{2}{15} t^3 (3t^2 - 5) + C$ Since $t = \sqrt{x+3}$, $t^2 = x+3$ and $t^3 = (x+3)\sqrt{x+3}$. $I = \frac{2}{15} (x+3)\sqrt{x+3} (3(x+3) - 5) + C$ $I = \frac{2}{15} \sqrt{x+3} (x+3) (3x + 9 - 5) + C$ $I = \frac{2}{15} \sqrt{x+3} (x+3) (3x + 4) + C$ $I = \frac{2}{15} \sqrt{x+3} (3x^2 + 4x + 9x + 12) + C$ $I = \frac{2}{15} \sqrt{x+3} (3x^2 + 13x + 12) + C$.

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

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