If int(2x+4)sqrt{x-1}dx = a(x-1)^{5/2} + b(x- | vedclass

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(B) Let $I = \int(2x+4)\sqrt{x-1}dx$. Substitute $t = x-1$,so $x = t+1$ and $dx = dt$. Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$. $I

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$\frac{28}{5}$

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(B) Let $I = \int(2x+4)\sqrt{x-1}dx$. Substitute $t = x-1$,so $x = t+1$ and $dx = dt$. Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$. $I = \int(2t^{3/2} + 6t^{1/2})dt$. Integrating term by term: $I = 2 \cdot \frac{t^{5/2}}{5/2} + 6 \cdot \frac{t^{3/2}}{3/2} + c$. $I = \frac{4}{5}t^{5/2} + 4t^{3/2} + c$. Substituting back $t = x-1$: $I = \frac{4}{5}(x-1)^{5/2} + 4(x-1)^{3/2} + c$. Comparing this with $a(x-1)^{5/2} + b(x-1)^{3/2} + c$,we get $a = \frac{4}{5}$ and $b = 4$. Therefore,$2a+b = 2(\frac{4}{5}) + 4 = \frac{8}{5} + 4 = \frac{8+20}{5} = \frac{28}{5}$.

If $\int(2x+4)\sqrt{x-1}dx = a(x-1)^{5/2} + b(x-1)^{3/2} + c$ where $c$ is a constant of integration,then the value of $(2a+b)$ is

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