Match list I with list II
| List - I | List - II | ||
| A. | \(\int\displaystyle\frac{dx}{x^2 - a^2} =\) | I. | \(\log_e\left|x + \sqrt{x^2 + a^2}\right| + C\) |
| B. | \(\int\displaystyle\frac{dx}{a^2 - x^2} =\) | II. | \(\displaystyle\frac{1}{2a}\log_e\left|\frac{x - a}{x + a}\right| + C\) |
| C. | \(\int\displaystyle\frac{dx}{\sqrt{x^2 - a^2}} =\) | III. | \(\displaystyle\frac{1}{2a}\log_e\left|\frac{a + x}{a - x}\right| + C\) |
| D. | \(\int\displaystyle\frac{dx}{\sqrt{x^2 + a^2}} =\) | IV. | \(\log_e\left|x + \sqrt{x^2 - a^2}\right| + C\) |
Choose the correct answer from the option given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-III, C-IV, D-I
All four are standard integrals. The first two come from partial fractions.
(A) \(\int\frac{dx}{x^2 - a^2}\)
This matches (II) \(\frac{1}{2a}\log_e\left|\frac{x - a}{x + a}\right| + C\).
\( \displaystyle \frac{1}{x^2 - a^2} = \frac{1}{2a}\left(\frac{1}{x - a} - \frac{1}{x + a}\right) \)
\( \displaystyle \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\left(\log_e|x - a| - \log_e|x + a|\right) + C \)
\( \displaystyle = \frac{1}{2a}\log_e\left|\frac{x - a}{x + a}\right| + C \)
(B) \(\int\frac{dx}{a^2 - x^2}\)
This matches (III) \(\frac{1}{2a}\log_e\left|\frac{a + x}{a - x}\right| + C\).
\( \displaystyle \frac{1}{a^2 - x^2} = \frac{1}{2a}\left(\frac{1}{a - x} + \frac{1}{a + x}\right) \)
\( \displaystyle \int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\left(-\log_e|a - x| + \log_e|a + x|\right) + C \)
\( \displaystyle = \frac{1}{2a}\log_e\left|\frac{a + x}{a - x}\right| + C \)
(C) \(\int\frac{dx}{\sqrt{x^2 - a^2}}\)
This matches (IV) \(\log_e\left|x + \sqrt{x^2 - a^2}\right| + C\), the standard result for \(\sqrt{x^2 - a^2}\) in the denominator.
(D) \(\int\frac{dx}{\sqrt{x^2 + a^2}}\)
This matches (I) \(\log_e\left|x + \sqrt{x^2 + a^2}\right| + C\), the standard result for \(\sqrt{x^2 + a^2}\) in the denominator.
So the matching is A-II, B-III, C-IV, D-I.
NCERT (2026-27 edition):
Integrals of Some Particular Functions
