Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{dx}{(x - 1)(x + 2)}\) | (I) | \(\displaystyle\frac{1}{3}\log|x - 1| + \frac{2}{3}\log|x + 2| + C\) |
| (B) | \(\int\displaystyle\frac{x}{(x - 1)(x + 2)}\,dx\) | (II) | \(\log(x^2 + 6x + 18) + C\) |
| (C) | \(\int\displaystyle\frac{dx}{x^2 + 6x + 18}\) | (III) | \(\displaystyle\frac{1}{3}\log\left|\frac{x - 1}{x + 2}\right| + C\) |
| (D) | \(\int\displaystyle\frac{2x + 6}{x^2 + 6x + 18}\,dx\) | (IV) | \(\displaystyle\frac{1}{3}\tan^{-1}\left(\frac{x + 3}{3}\right) + C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) \(\int \frac{dx}{(x - 1)(x + 2)}\)
This matches (III) \(\frac{1}{3}\log\left|\frac{x - 1}{x + 2}\right| + C\). In partial fractions:
\( \displaystyle \frac{1}{(x - 1)(x + 2)} = \frac{1}{3}\left(\frac{1}{x - 1} - \frac{1}{x + 2}\right) \)
\( \displaystyle \int \frac{dx}{(x - 1)(x + 2)} = \frac{1}{3}\left(\log|x - 1| - \log|x + 2|\right) + C \)
\( \displaystyle = \frac{1}{3}\log\left|\frac{x - 1}{x + 2}\right| + C \)
(B) \(\int \frac{x}{(x - 1)(x + 2)}\,dx\)
This matches (I) \(\frac{1}{3}\log|x - 1| + \frac{2}{3}\log|x + 2| + C\).
\( \displaystyle \frac{x}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} \)
Putting \(x = 1\): \(1 = 3A\), so \(A = \frac{1}{3}\).
Putting \(x = -2\): \(-2 = -3B\), so \(B = \frac{2}{3}\).
\( \displaystyle \int \frac{x}{(x - 1)(x + 2)}\,dx = \frac{1}{3}\log|x - 1| + \frac{2}{3}\log|x + 2| + C \)
(C) \(\int \frac{dx}{x^2 + 6x + 18}\)
This matches (IV) \(\frac{1}{3}\tan^{-1}\left(\frac{x + 3}{3}\right) + C\). Completing the square:
\( x^2 + 6x + 18 = (x + 3)^2 + 3^2 \)
With \(t = x + 3\) and \(\int \frac{dt}{t^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{t}{a} + C\), \(a = 3\):
\( \displaystyle \int \frac{dx}{(x + 3)^2 + 3^2} = \frac{1}{3}\tan^{-1}\left(\frac{x + 3}{3}\right) + C \)
(D) \(\int \frac{2x + 6}{x^2 + 6x + 18}\,dx\)
This matches (II) \(\log(x^2 + 6x + 18) + C\).
The numerator \(2x + 6\) is the derivative of the denominator, so the integral is \(\log|x^2 + 6x + 18| + C\). No modulus is needed because \(x^2 + 6x + 18 = (x + 3)^2 + 9 \gt 0\).