Match List-I with List-II

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

Choose the correct answer from the options given below:

Standard integral formulas used here (with $a = 4$ since $16 = 4^2$):

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

$\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1}\left(\dfrac{x}{a}\right) + C$

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

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

---

(A) $\int \dfrac{dx}{x^2 - 16}$

$= \dfrac{1}{2(4)} \log \left| \dfrac{x - 4}{x + 4} \right| + C$

$= \dfrac{1}{8} \log \left| \dfrac{x - 4}{x + 4} \right| + C$

Matches (III)

---

(B) $\int \dfrac{dx}{x^2 + 16}$

$= \dfrac{1}{4} \tan^{-1}\left(\dfrac{x}{4}\right) + C$

Matches (IV)

---

(C) $\int \dfrac{dx}{16 - x^2}$

$= \dfrac{1}{2(4)} \log \left| \dfrac{4 + x}{4 - x} \right| + C$

$= \dfrac{1}{8} \log \left| \dfrac{4 + x}{4 - x} \right| + C$

Matches (I)

---

(D) $\int \dfrac{dx}{\sqrt{x^2 - 16}}$

$= \log \left| x + \sqrt{x^2 - 16} \right| + C$

Matches (II)

---

Final matching:

(A) → (III)

(B) → (IV)

(C) → (I)

(D) → (II)

Therefore, the answer is Option 4.