The value of the integral $\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{{\left({(x+1)}^2{(1-x)}^6\right)}^{\frac{1}{4}}}dx$  is

$\begin{array}{l}\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{(1+\sqrt{3})}{{\left({(x+1)}^2{(1-x)}^6\right)}^{\frac{1}{4}}}dx=\underset{0}{\overset{\frac{1}{2}}{\int }}\begin{array}{l}\frac{(1+\sqrt{3})}{{\left({(1-x^2)}^2{(1-x)}^4\right)}^{\frac{1}{4}}}dx\\ \end{array}\\ =\left[\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1}{\sqrt{(1-x^2)}(1-x)}dx\right](1+\sqrt{3})\\ put\text{ x = sin }\theta \\ =\left[\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\cos \theta \text{d}\theta }{\cos \theta (1-\sin \theta )}\right](1+\sqrt{3})\\ =\left[\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{1+\sin \theta \text{ d}\theta }{{\cos }^2\theta }\right](1+\sqrt{3})\\ =\left[\underset{0}{\overset{\frac{\pi }{6}}{\int }}{\sec }^2\theta \text{d}\theta \text{+}\underset{0}{\overset{\frac{\pi }{6}}{\int }}\tan \theta \text{ sec}\theta \text{d}\theta \right](1+\sqrt{3})\\ =[{\tan \theta |}_0^{\frac{\pi }{6}}+\text{+ }{\sec \theta |}_0^{\frac{\pi }{6}}](1+\sqrt{3})\\ \frac{1}{\sqrt{3}}+(\frac{2}{\sqrt{3}}-1)(1+\sqrt{3})\\ =(\sqrt{3}-1)(\sqrt{3}+1)\\ ={\left(\sqrt{3}\right)}^2-1=2\end{array}$