int xsqrt{2x + 3} , dx = | vedclass

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Question

(A) To solve the integral $\int x\sqrt{2x + 3} \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x$,so $du = dx$. Let $

Vedclass · Accepted Answer

$\frac{x}{3}(2x + 3)^{3/2} - \frac{1}{15}(2x + 3)^{5/2} + c$

Vedclass · Answer

(A) To solve the integral $\int x\sqrt{2x + 3} \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x$,so $du = dx$. Let $dv = (2x + 3)^{1/2} \, dx$,so $v = \frac{(2x + 3)^{3/2}}{3/2 	imes 2} = \frac{1}{3}(2x + 3)^{3/2}$. Applying the formula: $\int x(2x + 3)^{1/2} \, dx = x \cdot \frac{1}{3}(2x + 3)^{3/2} - \int \frac{1}{3}(2x + 3)^{3/2} \, dx$. $= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{3} \cdot \frac{(2x + 3)^{5/2}}{5/2 	imes 2} + c$. $= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{3} \cdot \frac{(2x + 3)^{5/2}}{5} + c$. $= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{15}(2x + 3)^{5/2} + c$.

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

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