The value of the integral ∫01/2 1+3(x+1)2(1−x)61/4dx is__.

Solution:

$\begin{matrix} I = \int_{0}^{1 / 2} \frac{1 + \sqrt{3}}{{((x + 1)^{2} (1 - x)^{6})}^{1 / 4}} d x \\ = \int_{0}^{1 / 2} \frac{(1 + \sqrt{3}) d x}{(1 + x)^{2} {[\frac{(1 - x)^{6}}{(1 + x)^{6}}]}^{1 / 4}} \end{matrix}$

Put $\frac{1 - x}{1 + x} = t$

$\begin{array}{l} \Rightarrow \frac{- 2 d x}{(1 + x)^{2}} = d t \\ ∴ I = \int_{1}^{1 / 3} \frac{(1 + \sqrt{3}) d t}{- 2 t^{3 / 2}} = \frac{- (1 + \sqrt{3})}{2} \times {|\frac{- 2}{\sqrt{t}}|}_{1}^{1 / 3} \\ = (1 + \sqrt{3}) (\sqrt{3} - 1) = 2 \end{array}$

The value of the integral $\int_{0}^{1 / 2} \frac{1 + \sqrt{3}}{{((x + 1)^{2} (1 - x)^{6})}^{1 / 4}} d x$ is__.

Solution:

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