Step 1: Understanding the Concept:
The primary concept behind this matching problem is the identification of standard elementary integrals involving quadratic expressions.
These integrals form the foundation of integral calculus and are derived using various advanced integration techniques such as partial fractions and trigonometric substitutions.
Specifically, rational functions of the form \(1/(x^2 \pm a^2)\) generally result in logarithmic or inverse trigonometric functions.
In contrast, irrational functions containing square roots like \(\sqrt{x^2 \pm a^2}\) or \(\sqrt{a^2 - x^2}\) lead to inverse circular functions or more complex logarithmic forms.
Mastering these six to nine standard formulas is essential for solving any complex integration problem involving substitution or integration by parts.
Step 2: Key Formula or Approach:
To match these accurately, we use the following standard derivations:
1. For \(\int \frac{dx}{x^2 - a^2}\), the method of Partial Fractions is employed where we split the denominator into \((x-a)(x+a)\).
2. For \(\int \frac{dx}{a^2 - x^2}\), a similar partial fraction method is used, but the order of the terms in the resulting logarithm changes.
3. For \(\int \frac{dx}{\sqrt{x^2 - a^2}}\), we utilize the trigonometric substitution \(x = a \sec \theta\), which simplifies the integrand using the identity \(\sec^2 \theta - 1 = \tan^2 \theta\).
4. For \(\int \frac{dx}{\sqrt{a^2 - x^2}}\), we utilize the trigonometric substitution \(x = a \sin \theta\), simplifying via the identity \(1 - \sin^2 \theta = \cos^2 \theta\).
Step 3: Detailed Explanation:
Let's evaluate each part systematically to find the corresponding result in Column II.
Part A: Evaluate \( \int \frac{dx}{x^2 - a^2} \).
We decompose the integrand into partial fractions:
\[ \frac{1}{x^2 - a^2} = \frac{1}{(x-a)(x+a)} = \frac{1}{2a} \left( \frac{1}{x-a} - \frac{1}{x+a} \right) \]
Integrating both sides with respect to \(x\):
\[ \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \left( \int \frac{dx}{x-a} - \int \frac{dx}{x+a} \right) \]
\[ = \frac{1}{2a} (\log |x-a| - \log |x+a|) + C = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \]
This matches result III.
Part B: Evaluate \( \int \frac{dx}{a^2 - x^2} \).
Using a similar decomposition:
\[ \frac{1}{a^2 - x^2} = \frac{1}{(a-x)(a+x)} = \frac{1}{2a} \left( \frac{1}{a-x} + \frac{1}{a+x} \right) \]
Integrating both sides:
\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \left( \int \frac{dx}{a-x} + \int \frac{dx}{a+x} \right) \]
\[ = \frac{1}{2a} (-\log |a-x| + \log |a+x|) + C = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \]
This matches result I.
Part C: Evaluate \( \int \frac{dx}{\sqrt{x^2 - a^2}} \).
Substitute \(x = a \sec \theta\), so \(dx = a \sec \theta \tan \theta d\theta\).
The integral becomes:
\[ \int \frac{a \sec \theta \tan \theta d\theta}{\sqrt{a^2 \sec^2 \theta - a^2}} = \int \frac{a \sec \theta \tan \theta d\theta}{a \tan \theta} = \int \sec \theta d\theta \]
\[ = \log |\sec \theta + \tan \theta| + C' = \log \left| \frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a} \right| + C' \]
\[ = \log |x + \sqrt{x^2 - a^2}| - \log a + C' = \log |x + \sqrt{x^2 - a^2}| + C \]
This matches result II.
Part D: Evaluate \( \int \frac{dx}{\sqrt{a^2 - x^2}} \).
Substitute \(x = a \sin \theta\), so \(dx = a \cos \theta d\theta\).
The integral becomes:
\[ \int \frac{a \cos \theta d\theta}{\sqrt{a^2 - a^2 \sin^2 \theta}} = \int \frac{a \cos \theta d\theta}{a \cos \theta} = \int 1 d\theta = \theta + C \]
Since \(x = a \sin \theta\), we have \(\theta = \sin^{-1}(x/a)\).
Thus, the result is \(\sin^{-1} \frac{x}{a} + C\).
This matches result IV.
Step 4: Final Answer:
Matching all columns, we have: A $\rightarrow$ III, B $\rightarrow$ I, C $\rightarrow$ II, D $\rightarrow$ IV.
The corresponding option is (B).