When $T$ is a distribution and $k$ is a positive number, how should we define the rescaled distribution $T_k$? When $T$ is represented by a function $g$, we just want $g(kx)$. In terms of the integral against a test function $f$, this gives
$$
\int_{\mathbb{R}^n} g(kx) f(x) \,dx =k^{-n} \int_{\mathbb{R}^n} g(y) f(x/y) \,dx
$$
So, for a general distribution $T$ we define the rescaled distribution $T_k$ as
$$
T_k(f) = k^{-n} T(f(x/k))
$$
In the specific case of Dirac delta at $0$, the expression $k^{-n} T(f(x/k))$ evaluates to $ k^{-n}f(0)$. (You work in one dimension, $n=1$.)
One way of making sense of what $\int_a^b f(x) \delta(x) dx$ means is to see the Dirac delta as a measure. Now, I won't go into the details of what a measure is (if you are curious, head to the wikipedia page!). Let's just say that a measure is something that lets us make sense of the integral of real-valued functions over subsets. For a function $f: \mathbb{R} \to \mathbb{R}$, the integral of $f$ over the subset $A \subset \mathbb{R}$ with respect to the measure $\mu$ is usually written either
$
\int_{A} f \; d\mu
$
or
$
\int_{A} f(x) \; d\mu(x)
$
(if one wants to specify the variable of integration). Let me emphasize that it is very important that we specify the subset over which we are integrating!
The usual integral of functions over $\mathbb{R}$ is the integral with respect to Lebesgue measure, and the classical notation for the Lebesgue integral of a function $f$ over an interval with endpoints $a$ and $b$ is
$\int_{a}^{b} f(x) \; dx$. Now, notice that this last notation is ambiguous because it might mean either of the following integrals:
$$
\int_{[a,b)} f(x) \; dx \quad \text{or} \quad
\int_{(a,b)} f(x) \; dx \quad \text{or} \quad
\int_{(a,b]} f(x) \; dx \quad \text{or} \quad
\int_{[a,b]} f(x) \; dx
$$
However, in the case of Lebesgue measure, it turns out that all these integrals are equal and so the notation actually makes sense. The reason behind this is that the Lebesgue measure of a set is essentially its volume. Since points have zero volume, it is safe to ignore them in integrals.
Now, the point is that writing $\int_{a}^b f(x) \delta(x) dx$ is an abuse of notation, which can be dangerous if one is not careful. The Dirac delta is not a function on $\mathbb{R}$, so it does not make sense to integrate it with respect to the Lebesgue measure. What is really meant behind this notation is that $\int_{a}^b f(x) \delta(x) dx$ is the integral of $f$ with respect to the $\delta$ measure, namely it is the integral $\int_A f \; d\delta$, where $A$ is one of the intervals with endpoints $a$ and $b$.
Recall that in the case of Lebesgue measure, in didn't matter what choice of interval $A$ we made. For the Dirac measure, it is crucial that we specify over which interval we integrate. Here is why. Loosely speaking, the measure $\delta$ assigns "volume" $1$ to the set $\{ 0 \}$ and every set that contains $0$, and assigns "volume" $0$ to every other set, i.e. every set that doesn't contain $0$. This means that all the volume of the real line is concentrated at the point $0$. A consequence of this is that the integral of a function is given by:
$$
\int_A f \;d \delta =
\begin{cases}
f(0) & \text{if} \; 0 \in A\\
0 & \text{if} \; 0 \notin A
\end{cases}
$$
For most purposes, it is safe to think of this last equation as the definition of the Dirac delta. Now, notice that with this formula we have , for $a < 0$,
$$
\int_{(a, 0)} f \; d \delta = 0
$$
since $0 \notin (a, 0)$ while
$$
\int_{(a, 0]} f \; d \delta = f(0)
$$
since $0 \in ( a, 0]$. So, while it did not make any difference if we forgot about the point $0$ when computing the Lebesgue integral, it makes a huge difference when integrating with respect to the Dirac measure, because all the volume of the Dirac measure is concentrated at the point $0$.
To sum up, in order to make sense of the integral $\int_a ^b f(x) \delta(x) dx$, you should be careful to indicate what set you are integrating on. The notation $\int_A f(x) \; d\delta(x)$ can be a useful way of doing this.
Now to your next question: can we break up integrals in pieces?
Integrals with respect to measures share many properties with the usual Riemann integral (and in other ways they are very different, as I pointed out earlier). For example, the integral is linear. Another key property is that if a set $A$ can be written as a disjoint union of two sets, that is $A = B \cup C$ where $B \cap C = \emptyset$, then we can break up the integral:
$$
\int_A f \; d\mu = \int_B f \; d\mu + \int_C f \; d\mu.
$$
This means that you can safely break up integrals, as long as you keep track of the sets you are integrating on. What you should avoid at all cost is the following mistake:
$$
\int_a^b f(x) \delta(x) dx = \int_{a} ^0 f(x) \delta(x) dx + \int_0 ^b f(x) \delta(x) dx = f(0) + f(0)
$$
Best Answer
We will split the integral in two and make a substitution. $$\int_0^\infty f(x)\delta(x^2-9)\,dx=\int_0^\infty f(\sqrt x )\delta (x-9)\dfrac{1}{2x} \,dx=\dfrac{f(3)}{6}$$ Similarly, evaluating the other part of the integral and adding we get the answer.