Match List - I with List - II.
| 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) | \(\displaystyle\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C\) |
| (C) | \(\int\displaystyle\frac{dx}{\sqrt{x^2+a^2}}\) | (III) | \(\displaystyle\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C\) |
| (D) | \(\int\sqrt{x^2+a^2}\,dx\) | (IV) | \(\displaystyle\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
All four are standard integrals.
(A) By partial fractions, \(\displaystyle\frac{1}{x^2-a^2} = \frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)\), so
\( \displaystyle \int\frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C \quad \text{(II)} \)
(B) Here \(\displaystyle\frac{1}{a^2-x^2} = \frac{1}{2a}\left(\frac{1}{a-x}+\frac{1}{a+x}\right)\), so
\( \displaystyle \int\frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C \quad \text{(IV)} \)
(C) The standard result:
\( \displaystyle \int\frac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+C \quad \text{(I)} \)
(D) The standard result (integration by parts):
\( \displaystyle \int\sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C \quad \text{(III)} \)
So the match is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).
NCERT (2026-27 edition):
Integrals
