Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Functions | Integrals | ||
| (A) | \(\int\displaystyle\frac{dx}{x^2-4}, x \ne \pm 2\) | (I) | \(\log\left|x + \sqrt{4+x^2}\right|+C\), where C is an arbitrary constant |
| (B) | \(\int\displaystyle\frac{1}{\sqrt{16-x^2}}dx;\ |x| \lt 4\) | (II) | \(\sin^{-1}\left(\displaystyle\frac{x}{4}\right) +C\), where C is an arbitrary constant |
| (C) | \(\int\displaystyle\frac{1}{16+x^2}dx\) | (III) | \(\displaystyle\frac{1}{4}\log\left|\displaystyle\frac{x-2}{x+2}\right|+C\), where C is an arbitrary constant |
| (D) | \(\int\displaystyle\frac{1}{\sqrt{4+x^2}}dx\) | (IV) | \(\displaystyle\frac{1}{4}\tan^{-1}\left(\displaystyle\frac{x}{4}\right)+C\), where C is an arbitrary constant |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
Each integral is a standard integral.
(A) \(\int\frac{dx}{x^2-4}\)
This matches (III). Using \(\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C\) with a = 2:
\(\displaystyle \int\frac{dx}{x^2-4} = \frac{1}{4}\log\left|\frac{x-2}{x+2}\right| + C\)
(B) \(\int\frac{1}{\sqrt{16-x^2}}dx\)
This matches (II). Using \(\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C\) with a = 4:
\(\displaystyle \int\frac{dx}{\sqrt{16-x^2}} = \sin^{-1}\left(\frac{x}{4}\right) + C\)
(C) \(\int\frac{1}{16+x^2}dx\)
This matches (IV). Using \(\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\) with a = 4:
\(\displaystyle \int\frac{dx}{16+x^2} = \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) + C\)
(D) \(\int\frac{1}{\sqrt{4+x^2}}dx\)
This matches (I). Using \(\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C\) with a = 2:
\(\displaystyle \int\frac{dx}{\sqrt{4+x^2}} = \log\left|x + \sqrt{4+x^2}\right| + C\)
So the matching is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).