Match List-I with List-II (the constant of integration is left out in List-II)
| List-I (Integral) | List-II (Value) | ||
| (A) | \(\displaystyle\int \frac{dx}{x^2 - 1}\) | (I) | \(\log|x(x + 3)|\) |
| (B) | \(\displaystyle\int \frac{x\,dx}{(x + 1)(x + 2)}\) | (II) | \(\displaystyle\frac{1}{2}\log\left|\frac{x - 1}{x + 1}\right|\) |
| (C) | \(\displaystyle\int \frac{4\,dx}{(x - 1)(x + 3)}\) | (III) | \(\displaystyle\log\left|\frac{x - 1}{x + 3}\right|\) |
| (D) | \(\displaystyle\int \frac{(2x + 3)\,dx}{x(x + 3)}\) | (IV) | \(2\log|x + 2| - \log|x + 1|\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Each integrand is split into partial fractions over its two linear factors, and each piece integrates to a logarithm.
(A) \(\int \frac{dx}{x^2 - 1}\)
This matches (II) \(\frac{1}{2}\log\left|\frac{x - 1}{x + 1}\right|\). Here \(\frac{1}{(x - 1)(x + 1)} = \frac{1}{2}\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)\).
(B) \(\int \frac{x\,dx}{(x + 1)(x + 2)}\)
This matches (IV) \(2\log|x + 2| - \log|x + 1|\). Here \(\frac{x}{(x + 1)(x + 2)} = \frac{2}{x + 2} - \frac{1}{x + 1}\).
(C) \(\int \frac{4\,dx}{(x - 1)(x + 3)}\)
This matches (III) \(\log\left|\frac{x - 1}{x + 3}\right|\). Here \(\frac{4}{(x - 1)(x + 3)} = \frac{1}{x - 1} - \frac{1}{x + 3}\).
(D) \(\int \frac{(2x + 3)\,dx}{x(x + 3)}\)
This matches (I) \(\log|x(x + 3)|\). Here \(\frac{2x + 3}{x(x + 3)} = \frac{1}{x} + \frac{1}{x + 3}\).