Match List-I with List-II
Where C is arbitrary constant.| List-I | List-II | ||
| (A) | \( \int\displaystyle\frac{dx}{x^2-a^2}= \) | (I) | \( \log\left|x+\sqrt{x^2+a^2}\right|+C \) |
| (B) | \( \int\displaystyle\frac{dx}{a^2-x^2}= \) | (II) | \( \log\left|x+\sqrt{x^2-a^2}\right|+C \) |
| (C) | \( \int\displaystyle\frac{dx}{\sqrt{x^2-a^2}}= \) | (III) | \( \displaystyle\frac{1}{2a}\log\left|\displaystyle\frac{x-a}{x+a}\right|+C \) |
| (D) | \( \int\displaystyle\frac{dx}{\sqrt{x^2+a^2}}= \) | (IV) | \( \displaystyle\frac{1}{2a}\log\left|\displaystyle\frac{a+x}{a-x}\right|+C \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
These are standard integrals. The two without a root come from partial fractions; the two with a root have the form \( \log\left|x+\sqrt{\ \ }\right| \) with the same expression under the root.
(A) \( \int\frac{dx}{x^2-a^2} \)
This matches (III) \( \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C \).
\( \displaystyle \frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) \)
\( \displaystyle \int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C \)
(B) \( \int\frac{dx}{a^2-x^2} \)
This matches (IV) \( \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C \).
\( \displaystyle \frac{1}{a^2-x^2}=\frac{1}{2a}\left(\frac{1}{a-x}+\frac{1}{a+x}\right) \)
\( \displaystyle \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C \)
(C) \( \int\frac{dx}{\sqrt{x^2-a^2}} \)
This matches (II) \( \log\left|x+\sqrt{x^2-a^2}\right|+C \).
(D) \( \int\frac{dx}{\sqrt{x^2+a^2}} \)
This matches (I) \( \log\left|x+\sqrt{x^2+a^2}\right|+C \).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).
Tip: For \( \int\frac{dx}{x^2-a^2} \) and \( \int\frac{dx}{a^2-x^2} \), the term written first in the denominator (x or a) is the one that comes first in the log: \( x^2-a^2 \) gives \( \frac{x-a}{x+a} \), \( a^2-x^2 \) gives \( \frac{a+x}{a-x} \).
NCERT (2026-27 edition):
Integrals
