The value of the integral int_0^{frac{1}{2}} | vedclass

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Question

(C) Let $I = \int_0^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx$. Simplify the integrand: $((x+1)^2(1-x)^6)^{1/4} = (x+1)^{1/2}(1-x)^{3/2} = (1

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$2$

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(C) Let $I = \int_0^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx$. Simplify the integrand: $((x+1)^2(1-x)^6)^{1/4} = (x+1)^{1/2}(1-x)^{3/2} = (1-x)^2 \left(\frac{x+1}{1-x}\right)^{1/2}$. So,$I = \int_0^{1/2} \frac{1+\sqrt{3}}{\left(\frac{x+1}{1-x}\right)^{1/2} (1-x)^2} dx$. Let $t = \frac{x+1}{1-x}$. Then $dt = \frac{(1-x)(1) - (x+1)(-1)}{(1-x)^2} dx = \frac{1-x+x+1}{(1-x)^2} dx = \frac{2}{(1-x)^2} dx$. Thus,$\frac{dx}{(1-x)^2} = \frac{dt}{2}$. When $x=0$,$t=1$. When $x=1/2$,$t = \frac{1.5}{0.5} = 3$. $I = \int_1^3 \frac{1+\sqrt{3}}{\sqrt{t}} \cdot \frac{dt}{2} = \frac{1+\sqrt{3}}{2} \int_1^3 t^{-1/2} dt$. $I = \frac{1+\sqrt{3}}{2} [2\sqrt{t}]_1^3 = (1+\sqrt{3})(\sqrt{3}-1) = 3-1 = 2$.

The value of the integral $\int_0^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^2(1-x)^6\right)^{\frac{1}{4}}} d x$ is . . . . . . . .

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