Value of Definite Integral Involving Algebraic Radicals

To find the value of the given definite integral, we start with:

$I = \int_0^{\frac{1}{2}} \frac{1 + \sqrt{3}}{\left((x + 1)^2 (1 - x)^6\right)^{\frac{1}{4}}} dx$

First, simplify the expression in the denominator: $\left((x + 1)^2 (1 - x)^6\right)^{\frac{1}{4}} = (x + 1)^{\frac{2}{4}} (1 - x)^{\frac{6}{4}} = (x + 1)^{\frac{1}{2}} (1 - x)^{\frac{3}{2}}$

Rewriting the integral, we get: $I = (1 + \sqrt{3}) \int_0^{\frac{1}{2}} \frac{dx}{(x + 1)^{\frac{1}{2}} (1 - x)^{\frac{3}{2}}}$

We can rewrite the denominator in terms of the ratio $\frac{1 + x}{1 - x}$: $(x + 1)^{\frac{1}{2}} (1 - x)^{\frac{3}{2}} = (1 - x)^2 \cdot \sqrt{\frac{1 + x}{1 - x}}$

Thus, the integral becomes: $I = (1 + \sqrt{3}) \int_0^{\frac{1}{2}} \frac{dx}{(1 - x)^2 \sqrt{\frac{1 + x}{1 - x}}}$

Now, use the substitution: $t = \frac{1 + x}{1 - x}$

Differentiating both sides with respect to $x$: $dt = \frac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^2} dx = \frac{2}{(1 - x)^2} dx \implies \frac{dx}{(1 - x)^2} = \frac{dt}{2}$

Next, change the limits of integration:

• When $x = 0$, $t = \frac{1 + 0}{1 - 0} = 1$
• When $x = \frac{1}{2}$, $t = \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = \frac{3/2}{1/2} = 3$

Substitute these into the integral: $I = (1 + \sqrt{3}) \int_1^3 \frac{dt}{2 \sqrt{t}}$

$I = \frac{1 + \sqrt{3}}{2} \int_1^3 t^{-\frac{1}{2}} dt$

$I = \frac{1 + \sqrt{3}}{2} \left[ 2 \sqrt{t} \right]_1^3$

$I = (1 + \sqrt{3}) (\sqrt{3} - 1)$

Using the difference of squares formula, $(a+b)(a-b) = a^2 - b^2$: $I = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2$

Final Answer: The value of the definite integral is 2.