Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\displaystyle\int\frac{dx}{x^2-16}\) | (I) | \(\displaystyle\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c\), where C is an arbitrary constant |
| (B) | \(\displaystyle\int\frac{dx}{x^2+16}\) | (II) | \(\log\left|x + \sqrt{x^2-16}\right|+c\), where C is an arbitrary constant |
| (C) | \(\displaystyle\int\frac{dx}{16-x^2}\) | (III) | \(\displaystyle\frac{1}{8}\log\left|\frac{x-4}{x+4}\right|+c\), where C is an arbitrary constant |
| (D) | \(\displaystyle\int\frac{dx}{\sqrt{x^2-16}}\) | (IV) | \(\displaystyle\frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right)+c\), where C is an arbitrary constant |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each integral is a standard integral with a = 4, so \(\frac{1}{2a} = \frac{1}{8}\) and \(\frac{1}{a} = \frac{1}{4}\).
(A) \(\displaystyle\int\frac{dx}{x^2-16}\)
This matches (III). Using \(\int\frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C\):
\( \displaystyle \int\frac{dx}{x^2-16} = \frac{1}{8}\log\left|\frac{x-4}{x+4}\right| + C \)
(B) \(\displaystyle\int\frac{dx}{x^2+16}\)
This matches (IV). Using \(\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\):
\( \displaystyle \int\frac{dx}{x^2+16} = \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) + C \)
(C) \(\displaystyle\int\frac{dx}{16-x^2}\)
This matches (I). Using \(\int\frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + C\):
\( \displaystyle \int\frac{dx}{16-x^2} = \frac{1}{8}\log\left|\frac{4+x}{4-x}\right| + C \)
(D) \(\displaystyle\int\frac{dx}{\sqrt{x^2-16}}\)
This matches (II). Using \(\int\frac{dx}{\sqrt{x^2-a^2}} = \log\left|x + \sqrt{x^2-a^2}\right| + C\):
\( \displaystyle \int\frac{dx}{\sqrt{x^2-16}} = \log\left|x + \sqrt{x^2-16}\right| + C \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
NCERT (2026-27 edition):
Integrals
