Q87 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
$\int \dfrac{x + 2}{x^2 - 7x + 12} dx$ is equal to
- Correct answerA$-5\log|x - 3| + 6\log|x - 4| + c$
- B$-5\log|x - 3| - 6\log|x - 4| + c$
- C$5\log|x - 3| - 6\log|x - 4| + c$
- D$5\log|x - 3| + 6\log|x - 4| + c$
Solution
Let $I = \int \dfrac{x + 2}{x^2 - 7x + 12} dx$
Factorizing the denominator, we get $x^2 - 7x + 12 = (x - 3)(x - 4)$
Using partial fractions:
$\dfrac{x + 2}{(x - 3)(x - 4)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 4}$
$x + 2 = A(x - 4) + B(x - 3)$
Substituting $x = 3$, we get $5 = -A \Rightarrow A = -5$
Substituting $x = 4$, we get $6 = B \Rightarrow B = 6$
Therefore, the integral becomes:
$I = \int \left( \dfrac{-5}{x - 3} + \dfrac{6}{x - 4} \right) dx$
$I = -5\log|x - 3| + 6\log|x - 4| + c$
Answer: $-5\log|x - 3| + 6\log|x - 4| + c$
Factorizing the denominator, we get $x^2 - 7x + 12 = (x - 3)(x - 4)$
Using partial fractions:
$\dfrac{x + 2}{(x - 3)(x - 4)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 4}$
$x + 2 = A(x - 4) + B(x - 3)$
Substituting $x = 3$, we get $5 = -A \Rightarrow A = -5$
Substituting $x = 4$, we get $6 = B \Rightarrow B = 6$
Therefore, the integral becomes:
$I = \int \left( \dfrac{-5}{x - 3} + \dfrac{6}{x - 4} \right) dx$
$I = -5\log|x - 3| + 6\log|x - 4| + c$
Answer: $-5\log|x - 3| + 6\log|x - 4| + c$
Hence, the correct answer is option: 'A'.