Match List I with List II

List I	List II
A. $\int \frac{dx}{x^2 - a^2}$ is equal to	I. $\frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$
B. $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is equal to	II. $\frac{1}{2a}\log\left\lvert \frac{x-a}{x+a}\right\rvert + C$
C. $\int \sqrt{a^2 - x^2} \, dx$ is equal to	III. $\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$
D. $\int \sqrt{a^2 + x^2} \, dx$ is equal to	IV. $\log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$

Choose the correct answer from the options given below:

We need to match four standard integral formulas with their correct results.

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$\int \dfrac{dx}{x^2 - a^2}$

Using partial fractions, we write $\dfrac{1}{x^2 - a^2} = \dfrac{1}{(x-a)(x+a)}$, which after integrating gives:

$\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log\left\lvert \dfrac{x-a}{x+a}\right\rvert + C$

This matches $\text{II}$.

So, $A \to II$

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$\int \dfrac{dx}{\sqrt{x^2 + a^2}}$

Substituting $x = a\tan\theta$ and simplifying, we get:

$\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$

This matches $\text{IV}$.

So, $B \to IV$

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$\int \sqrt{a^2 - x^2}\, dx$

Substituting $x = a\sin\theta$ and using the identity $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$, we get:

$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$

The $\sin^{-1}$ term appears because the integrand contains $\sqrt{a^2 - x^2}$ (minus sign inside the root).

This matches $\text{III}$.

So, $C \to III$

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$\int \sqrt{a^2 + x^2}\, dx$

Substituting $x = a\tan\theta$ and simplifying, we get:

$\int \sqrt{a^2 + x^2}\, dx = \dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$

Since there is a plus sign inside the root, the result has a $\log$ term instead of $\sin^{-1}$.

This matches $\text{I}$.

So, $D \to I$

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$A \to II, \quad B \to IV, \quad C \to III, \quad D \to I$