Match List-I with List-II List-I List-II (A) ∫dx/(x^2-16) (I) 1/8log|…

Match List-I with List-II

List-I	List-II
(A) $\int\frac{dx}{x^2-16}$	(I) $\frac{1}{8}\log\|\frac{4+x}{4-x}\|+c$, Where C is an arbitrary constant,
(B) $\int\frac{dx}{x^2+16}$	(II) $\log\|x + \sqrt{x^2-16}\|+c$, Where C is an arbitrary constant,
(C) $\int\frac{dx}{16-x^2}$	(III) $\frac{1}{8}\log\|\frac{x-4}{x+4}\|+c$, Where C is an arbitrary constant,
(D) $\int\frac{dx}{\sqrt{x^2-16}}$	(IV) $\frac{1}{4}\tan^{-1}(\frac{x}{4})+c$, Where C is an arbitrary constant,

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

List-I	List-II
(A) $\int\frac{dx}{x^2-16}$	(III) $\frac{1}{8}\log\|\frac{x-4}{x+4}\|+c$, Where C is an arbitrary constant,
(B) $\int\frac{dx}{x^2+16}$	(IV) $\frac{1}{4}\tan^{-1}(\frac{x}{4})+c$, Where C is an arbitrary constant,
(C) $\int\frac{dx}{16-x^2}$	(I) $\frac{1}{8}\log\|\frac{4+x}{4-x}\|+c$, Where C is an arbitrary constant,
(D) $\int\frac{dx}{\sqrt{x^2-16}}$	(II) $\log\|x + \sqrt{x^2-16}\|+c$, Where C is an arbitrary constant,

Use: $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C$, $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C$

Use: $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)+C$, $\int \frac{dx}{\sqrt{x^2-a^2}}=\log\!\left|x+\sqrt{x^2-a^2}\right|+C$

(A) $\int \frac{dx}{x^2-16}\ \to\ \frac{1}{8}\log\left|\frac{x-4}{x+4}\right|$ ⇒ (III)

(B) $\int \frac{dx}{x^2+16}\ \to\ \frac{1}{4}\tan^{-1}\!\left(\frac{x}{4}\right)$ ⇒ (IV)

(D) $\int \frac{dx}{\sqrt{x^2-16}}\ \to\ \log\!\left|x+\sqrt{x^2-16}\right|$ ⇒ (II)

Matching: A–III, B–IV, C–I, D–II