Match List-I with List-II (\(-2 \lt x \lt 2\), C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{dx}{4-x^2}\) | (I) | \(-\log|2-x|+C\) |
| (B) | \(\int\displaystyle\frac{x}{4-x^2}\,dx\) | (II) | \(\displaystyle\frac{1}{4}\log\left|\frac{2+x}{2-x}\right|+C\) |
| (C) | \(\int\displaystyle\frac{2-x}{4-x^2}\,dx\) | (III) | \(\log|2+x|+C\) |
| (D) | \(\int\displaystyle\frac{2+x}{4-x^2}\,dx\) | (IV) | \(\displaystyle -\frac{1}{2}\log\left|4-x^2\right|+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
All four have the denominator \(4-x^2 = (2-x)(2+x)\); the numerator decides the method.
(A) \(\int\frac{dx}{4-x^2}\)
This matches (II) \(\frac{1}{4}\log\left|\frac{2+x}{2-x}\right|+C\).
Using the standard integral \(\int\frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|\) with a = 2:
\( \displaystyle \int\frac{dx}{4-x^2} = \frac{1}{4}\log\left|\frac{2+x}{2-x}\right|+C \)
(B) \(\int\frac{x}{4-x^2}\,dx\)
This matches (IV) \(-\frac{1}{2}\log\left|4-x^2\right|+C\).
The numerator is \(-\frac{1}{2}\) times the derivative \(-2x\) of the denominator:
\( \displaystyle \int\frac{x}{4-x^2}\,dx = -\frac{1}{2}\int\frac{-2x}{4-x^2}\,dx = -\frac{1}{2}\log\left|4-x^2\right|+C \)
(C) \(\int\frac{2-x}{4-x^2}\,dx\)
This matches (III) \(\log|2+x|+C\).
The factor \(2-x\) cancels:
\( \displaystyle \int\frac{2-x}{(2-x)(2+x)}\,dx = \int\frac{dx}{2+x} = \log|2+x|+C \)
(D) \(\int\frac{2+x}{4-x^2}\,dx\)
This matches (I) \(-\log|2-x|+C\).
The factor \(2+x\) cancels, and \(\int\frac{dx}{2-x}\) carries a minus sign because the derivative of \(2-x\) is \(-1\):
\( \displaystyle \int\frac{dx}{2-x} = -\log|2-x|+C \)