The value of the integral $\int_{0}^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{\frac{1}{4}}}\,dx$ is ______.

The value of the integral ∫₀^(1/2) (1+√3)/(((x+1)²(1-x)⁶)^(1/4)) dx is ______. $\displaystyle\int_{0}^{1/2} \dfrac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{1/4}}\,dx$ $\text{To find: the value of the integral}$ $1.\ \text{simplify the denominator}$ $2.\ \text{spot the antiderivative}$ $3.\ \text{apply the limits } 0 \to 1/2$ $\left((x+1)^{2}(1-x)^{6}\right)^{1/4} = (x+1)^{1/2}(1-x)^{3/2}$ $I = (1+\sqrt{3})\displaystyle\int_{0}^{1/2} (x+1)^{-1/2}(1-x)^{-3/2}\,dx$ $\text{Find } y \text{ with } y' = (x+1)^{-1/2}(1-x)^{-3/2}$ $y = \sqrt{\dfrac{x+1}{1-x}} = (x+1)^{1/2}(1-x)^{-1/2}$ $y' = \dfrac{1}{2}(x+1)^{-1/2}(1-x)^{-1/2} + \dfrac{1}{2}(x+1)^{1/2}(1-x)^{-3/2}$ $= \dfrac{1}{2}(x+1)^{-1/2}(1-x)^{-3/2}\left[(1-x)+(x+1)\right] = (x+1)^{-1/2}(1-x)^{-3/2}$ $I = (1+\sqrt{3})\left[\sqrt{\dfrac{x+1}{1-x}}\right]_{0}^{1/2}$ $\text{At } x = 1/2:\quad \sqrt{\dfrac{1/2+1}{1-1/2}} = \sqrt{3}$ $\text{At } x = 0:\quad \sqrt{\dfrac{0+1}{1-0}} = 1$ $I = (1+\sqrt{3})(\sqrt{3} - 1)$ $= (\sqrt{3})^{2} - 1^{2} = 3 - 1$ $I = 2$ $\boxed{\text{Answer: } 2}$