Match List-I with List-II List-I Functions List-II Integers (A) ∫dx/(…

The correct answer is Option (2) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

List-I Functions	List-II Integers
(A) $\int\frac{dx}{x^2-4},x≠±2$	(III) $\frac{1}{4}\log\left|\frac{x-2}{x+2}\right|+C$, where C is an arbitary constant
(B) $\int\frac{1}{\sqrt{16-x^2}}dx;|x|<4$	(II) $\sin^{-1}(\frac{x}{4}) +C$, where C is an arbitary constant
(C) $\int\frac{1}{16+x^2}dx$	(IV) $\frac{1}{4}\tan^{-1}(\frac{x}{4})+C$, where C is an arbitary constant
(D) $\int\frac{1}{\sqrt{4+x^2}}dx$	(I) $\log\left|x + \sqrt{4+x^2}\right|+C$, where C is an arbitary constant

Explanation:

(A) $\displaystyle \int \frac{dx}{x^{2}-4}$

Use the standard result $\displaystyle \int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C$.

Here $a=2$, so the answer is $\displaystyle \frac{1}{4}\log\left|\frac{x-2}{x+2}\right| + C$, which matches (III).

(B) $\displaystyle \int \frac{1}{\sqrt{16-x^{2}}}\,dx$

Use the standard result $\displaystyle \int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}\left(\frac{x}{a}\right) + C$.

Here $a=4$, so the answer is $\displaystyle \sin^{-1}\left(\frac{x}{4}\right) + C$, which matches (II).

(C) $\displaystyle \int \frac{1}{16+x^{2}}\,dx$

Use the standard result $\displaystyle \int \frac{dx}{a^{2}+x^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$.

Here $a=4$, so the answer is $\displaystyle \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) + C$, which matches (IV).

(D) $\displaystyle \int \frac{1}{\sqrt{4+x^{2}}}\,dx$

Use the standard result $\displaystyle \int \frac{dx}{\sqrt{x^{2}+a^{2}}} = \log\left|x+\sqrt{x^{2}+a^{2}}\right| + C$.

Here $a=2$, so the answer is $\displaystyle \log\left|x+\sqrt{4+x^{2}}\right| + C$, which matches (I).