Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{dx}{x^2+4x}\) | (I) | \(\displaystyle\frac{1}{2}\log\left|x^2+4x\right|+C\) |
| (B) | \(\int\displaystyle\frac{x+2}{x^2+4x}\,dx\) | (II) | \(\displaystyle\frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C\) |
| (C) | \(\int\displaystyle\frac{dx}{x^2+4x+4}\) | (III) | \(\displaystyle\frac{1}{4}\log\left|\frac{x}{x+4}\right|+C\) |
| (D) | \(\int\displaystyle\frac{dx}{x^2+4x+8}\) | (IV) | \(\displaystyle -\frac{1}{x+2}+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The four denominators differ only in the constant, and the nature of the quadratic decides the method.
(A) \(\int\frac{dx}{x^2+4x}\)
This matches (III) \(\frac{1}{4}\log\left|\frac{x}{x+4}\right|+C\).
The denominator has two linear factors, so by partial fractions:
\( \displaystyle \frac{1}{x(x+4)} = \frac{1}{4}\left(\frac{1}{x}-\frac{1}{x+4}\right) \)
(B) \(\int\frac{x+2}{x^2+4x}\,dx\)
This matches (I) \(\frac{1}{2}\log\left|x^2+4x\right|+C\).
The numerator is half the derivative \(2x+4\) of the denominator:
\( \displaystyle \int\frac{x+2}{x^2+4x}\,dx = \frac{1}{2}\int\frac{2x+4}{x^2+4x}\,dx = \frac{1}{2}\log\left|x^2+4x\right|+C \)
(C) \(\int\frac{dx}{x^2+4x+4}\)
This matches (IV) \(-\frac{1}{x+2}+C\).
The denominator is a perfect square:
\( \displaystyle \int\frac{dx}{(x+2)^2} = -\frac{1}{x+2}+C \)
(D) \(\int\frac{dx}{x^2+4x+8}\)
This matches (II) \(\frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C\).
Completing the square, \(x^2+4x+8 = (x+2)^2+2^2\):
\( \displaystyle \int\frac{dx}{(x+2)^2+2^2} = \frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C \)