$\displaystyle\int \dfrac{\left(\sqrt[3]{x + \sqrt{2 - x^2}}\right)\left(\sqrt[6]{1 - x\sqrt{2 - x^2}}\right)}{\sqrt[3]{1 - x^2}}\,dx; \, (x \in (0, 1)) =$

Let the given integral be $I = \int \dfrac{(x + \sqrt{2 - x^2})^{1/3} (1 - x\sqrt{2 - x^2})^{1/6}}{(1 - x^2)^{1/3}} \, dx$.

Consider the first term in the numerator. We can rewrite it by squaring the expression inside and changing the exponent:

$(x + \sqrt{2 - x^2})^{1/3} = \left((x + \sqrt{2 - x^2})^2\right)^{1/6}$

Expanding the square, we get:

$(x + \sqrt{2 - x^2})^2 = x^2 + (2 - x^2) + 2x\sqrt{2 - x^2} = 2 + 2x\sqrt{2 - x^2} = 2(1 + x\sqrt{2 - x^2})$

Substituting this back, the first term becomes:

$(x + \sqrt{2 - x^2})^{1/3} = 2^{1/6} (1 + x\sqrt{2 - x^2})^{1/6}$

Now, multiply this with the second term in the numerator:

Numerator $= 2^{1/6} (1 + x\sqrt{2 - x^2})^{1/6} (1 - x\sqrt{2 - x^2})^{1/6}$

Using the identity $(a+b)(a-b) = a^2 - b^2$, we get:

Numerator $= 2^{1/6} \left(1 - x^2(2 - x^2)\right)^{1/6}$

Numerator $= 2^{1/6} (1 - 2x^2 + x^4)^{1/6}$

Numerator $= 2^{1/6} \left((1 - x^2)^2\right)^{1/6}$

Since $x \in (0, 1)$, $1 - x^2 > 0$, so this simplifies to:

Numerator $= 2^{1/6} (1 - x^2)^{1/3}$

Substituting the simplified numerator back into the integral:

$I = \int \dfrac{2^{1/6} (1 - x^2)^{1/3}}{(1 - x^2)^{1/3}} \, dx$

$I = \int 2^{1/6} \, dx$

$I = 2^{1/6}x + c$

Answer: $2^{\dfrac{1}{6}}x + c$