If C is an arbitrary constant, then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( \int a^{x}\,dx \) | I. | \( \sin^{-1}\displaystyle\frac{x}{a}+C \) |
| B. | \( \int\displaystyle\frac{dx}{\sqrt{a^{2}-x^{2}}} \) | II. | \( \log_{e}\left|x+\sqrt{x^{2}-a^{2}}\right|+C \) |
| C. | \( \int\displaystyle\frac{dx}{\sqrt{a^{2}+x^{2}}} \) | III. | \( \displaystyle\frac{a^{x}}{\log_{e}a}+C \) |
| D. | \( \int\displaystyle\frac{dx}{\sqrt{x^{2}-a^{2}}} \) | IV. | \( \log_{e}\left|x+\sqrt{a^{2}+x^{2}}\right|+C \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-IV, D-II
All four are standard integrals.
(A) \( \int a^{x}\,dx \)
This matches (III) \( \frac{a^{x}}{\log_{e}a}+C \), since \( \frac{d}{dx}a^{x}=a^{x}\log_{e}a \).
(B) \( \int\frac{dx}{\sqrt{a^{2}-x^{2}}} \)
This matches (I) \( \sin^{-1}\frac{x}{a}+C \).
(C) \( \int\frac{dx}{\sqrt{a^{2}+x^{2}}} \)
This matches (IV) \( \log_{e}\left|x+\sqrt{a^{2}+x^{2}}\right|+C \). Differentiating confirms it:
\( \displaystyle \frac{d}{dx}\log\left(x+\sqrt{x^{2}+a^{2}}\right)=\frac{1+\frac{x}{\sqrt{x^{2}+a^{2}}}}{x+\sqrt{x^{2}+a^{2}}} \)
\( \displaystyle =\frac{1}{\sqrt{x^{2}+a^{2}}} \)
(D) \( \int\frac{dx}{\sqrt{x^{2}-a^{2}}} \)
This matches (II) \( \log_{e}\left|x+\sqrt{x^{2}-a^{2}}\right|+C \), by the same working with \( -a^{2} \) in place of \( a^{2} \).
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).