integral of delta function of x^2

$\delta(f(x))$ is only defined if $\nabla f \neq 0$ wherever $f(x) = 0$. In this case, yes it would be "infinite", which is why it is not a well defined object.

Generally we assign the interpretation that

$$\delta(f(x)) \equiv \sum_{f(x_i)=0} \frac{1}{|\nabla f(x_i)|}\delta(x-x_i)$$

The problem here is a problem of definition.

First of all, by definition the dirac delta function is (as a distribution with finite support) a linear form defined on $\mathcal{C}^0(\mathbb{R})$ that to each function $\phi$ associates $$\int_{-\infty}^\infty\delta(x)\phi(x)dx:=\phi(0).$$

Please note that there is no definition of $\delta(x^2)$ other than by the means of change of variable !

To test what $\delta(x^2)$ should be you need to apply everything to a test function :

$$\int_{-\infty}^{+\infty}\delta(x^2)\phi(x)dx=\int_{-\infty}^\infty\delta(y) \frac{\phi(\sqrt{y})}{\sqrt{y}} 1_{y\geq 0} dy$$

BUT the function you wish to test $\delta$ against is not in general continuous at the point $0$, even for $\phi$ smooth with compact support so this makes no sense.