Concept:
This problem is based on standard integrals involving algebraic expressions of the forms:
\[
x^2-a^2,\qquad a^2-x^2,\qquad \sqrt{x^2-a^2},\qquad \sqrt{a^2-x^2}
\]
These are extremely important standard results in Integral Calculus and are frequently used in competitive examinations such as CUET, JEE Main, and JEE Advanced.
The important standard integrals used in this question are:
\[
\int \frac{dx}{x^2-a^2}
=
\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
\]
\[
\int \frac{dx}{a^2-x^2}
=
\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
\]
\[
\int \frac{dx}{\sqrt{x^2-a^2}}
=
\log\left|x+\sqrt{x^2-a^2}\right|+C
\]
\[
\int \frac{dx}{\sqrt{a^2-x^2}}
=
\sin^{-1}\left(\frac{x}{a}\right)+C
\]
We now match each integral carefully one by one.
Step 1: Matching Integral A
We are given:
\[
A=\int \frac{dx}{x^2-a^2}
\]
Recall the standard result:
\[
\int \frac{dx}{x^2-a^2}
=
\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
\]
This exactly matches expression III.
Therefore,
\[
A \longrightarrow III
\]
Step 2: Matching Integral B
We are given:
\[
B=\int \frac{dx}{a^2-x^2}
\]
Using the standard formula:
\[
\int \frac{dx}{a^2-x^2}
=
\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
\]
This matches expression I.
Hence,
\[
B \longrightarrow I
\]
Step 3: Matching Integral C
We are given:
\[
C=\int \frac{dx}{\sqrt{x^2-a^2}}
\]
Using the standard integral:
\[
\int \frac{dx}{\sqrt{x^2-a^2}}
=
\log\left|x+\sqrt{x^2-a^2}\right|+C
\]
This corresponds exactly to expression II.
Therefore,
\[
C \longrightarrow II
\]
Step 4: Matching Integral D
We are given:
\[
D=\int \frac{dx}{\sqrt{a^2-x^2}}
\]
Using the standard trigonometric integral formula:
\[
\int \frac{dx}{\sqrt{a^2-x^2}}
=
\sin^{-1}\left(\frac{x}{a}\right)+C
\]
This matches expression IV.
Hence,
\[
D \longrightarrow IV
\]
Step 5: Writing the final matching
Collecting all the obtained matches:
\[
A \rightarrow III
\]
\[
B \rightarrow I
\]
\[
C \rightarrow II
\]
\[
D \rightarrow IV
\]
Thus the correct option is:
\[
\boxed{(B)\ A-III,\ B-I,\ C-II,\ D-IV}
\]