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

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Question

(C) Let $I = \int(2x+4)\sqrt{x-1} \, dx$. Substitute $x-1 = t^2$,which implies $x = t^2+1$ and $dx = 2t \, dt$. Substituting these into the integral:

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$\frac{24}{5}$

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(C) Let $I = \int(2x+4)\sqrt{x-1} \, dx$. Substitute $x-1 = t^2$,which implies $x = t^2+1$ and $dx = 2t \, dt$. Substituting these into the integral: $I = \int(2(t^2+1)+4) \cdot t \cdot (2t) \, dt$ $I = \int(2t^2+2+4) \cdot 2t^2 \, dt$ $I = \int(2t^2+6) \cdot 2t^2 \, dt$ $I = \int(4t^4 + 12t^2) \, dt$ Integrating term by term: $I = 4 \int t^4 \, dt + 12 \int t^2 \, dt$ $I = 4 \cdot \frac{t^5}{5} + 12 \cdot \frac{t^3}{3} + c$ $I = \frac{4}{5}t^5 + 4t^3 + c$ Substituting back $t = (x-1)^{\frac{1}{2}}$: $I = \frac{4}{5}(x-1)^{\frac{5}{2}} + 4(x-1)^{\frac{3}{2}} + c$ Comparing this with the given form $a(x-1)^{\frac{5}{2}} + b(x-1)^{\frac{3}{2}} + c$,we get $a = \frac{4}{5}$ and $b = 4$. Therefore,$a+b = \frac{4}{5} + 4 = \frac{4+20}{5} = \frac{24}{5}$.

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

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