Animated Solution for Mathematics - Definite Integration: The integral $\frac{24}{\pi }{\int }_0^{\sqrt{2}}\frac{(2-x^2)dx}{(2+x^2)\sqrt{4+x^4}}$ is equal to ______.

The given integral is:

At first glance, this expression appears daunting. However, in the context of JEE Advanced, such complexity is often a mask for underlying symmetry.

The first step is to observe the powers of $x$. We have $x^0$ (in the constant 2), $x^2$, and $x^4$.

Since every power is even, we identify a 'Symmetry Signature.' This suggests that dividing the numerator and denominator by $x^2$ will transform the expression into a form involving $(x+\frac{k}{x})$, which is the key to substitution.

We divide the numerator and the denominator by $x^2$. The numerator becomes:

For the denominator, we split the $x^2$ into $x\cdot x$. We divide the first part $(2+x^2)$ by $x$ to get $(\frac{2}{x}+x)$, and we push the second $x$ inside the square root as $x^2$:

The integral is now transformed into:

We define our substitution variable as $t=x+\frac{2}{x}$. Differentiating this yields $dt=(1-\frac{2}{x^2})dx$, which implies that the numerator $(\frac{2}{x^2}-1)dx$ is exactly $-dt$.

To handle the square root, we note that:

Substituting these into the integral, we obtain:

We must update the limits of integration. As $x\to 0^{+}$, $t=x+\frac{2}{x}\to \infty$. As $x\to \sqrt{2}$, $t=\sqrt{2}+\frac{2}{\sqrt{2}}=2\sqrt{2}$.

Using the negative sign to flip the limits, the integral becomes:

We utilize the standard integral form $\int \frac{dt}{t\sqrt{t^2-a^2}}=\frac{1}{a}{\sec }^{-1}(\frac{t}{a})$, where $a=2$. Evaluating this from $2\sqrt{2}$ to $\infty$:

The final result is 3.