1. Rewrite the integrand

Given $I=\int_0^{1/2} \frac{1+\sqrt3}{\left((x+1)^2(1-x)^6\right)^{1/4}}\,dx$

First simplify the denominator: $\left((x+1)^2(1-x)^6\right)^{1/4}=(x+1)^{1/2}(1-x)^{3/2}$ for $0\le x\le \tfrac12$.

So, $I=(1+\sqrt3)\int_0^{1/2}\frac{dx}{\sqrt{x+1}(1-x)^{3/2}}$

2. Look for a useful substitution

Let $t=\sqrt{\frac{x+1}{1-x}}$ Then $t^2=\frac{x+1}{1-x}$

Differentiate implicitly: $2t\,\frac{dt}{dx}=\frac{(1-x)-(-(x+1))}{(1-x)^2}=\frac{2}{(1-x)^2}$ Thus, $\frac{dt}{dx}=\frac{1}{t(1-x)^2}$

Now since $t=\frac{\sqrt{x+1}}{\sqrt{1-x}},$ we get $t(1-x)^2=\sqrt{x+1}(1-x)^{3/2}$ Hence $dt=\frac{dx}{\sqrt{x+1}(1-x)^{3/2}}$

So the integral becomes simply $I=(1+\sqrt3)\int dt$

3. Change the limits

When $x=0$, $t=\sqrt{\frac{1}{1}}=1$

When $x=\frac12$, $t=\sqrt{\frac{\frac32}{\frac12}}=\sqrt3$

Therefore, $I=(1+\sqrt3)\int_1^{\sqrt3} dt=(1+\sqrt3)(\sqrt3-1)$

4. Simplify

$(1+\sqrt3)(\sqrt3-1)=3-1=2$

So, $\boxed{2}$

5. Comparison with stored correct answer

Derived answer = $2$

Stored correct answer = $2$

They match.