Q88 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics

$\int \dfrac{2x}{4 - 3x - x^2} dx = \cdots$

  • A
    $\dfrac{8}{5}\log|4 + x| - \dfrac{2}{5}\log|1 - x| + c$
  • B
    $2\log|4 + x| + \dfrac{2}{5}\log|1 - x| + c$
  • C
    $-\log|4 + x| + \dfrac{2}{5}\log|1 - x| + c$
  • Correct answer
    D
    $-\dfrac{8}{5}\log|4 + x| - \dfrac{2}{5}\log|1 - x| + c$

Solution

Let $I = \int \dfrac{2x}{4 - 3x - x^2} dx$

Factorizing the denominator:

$4 - 3x - x^2 = (1 - x)(4 + x)$

Using partial fractions:

$\dfrac{2x}{(1 - x)(4 + x)} = \dfrac{A}{1 - x} + \dfrac{B}{4 + x}$

$2x = A(4 + x) + B(1 - x)$

Substituting $x = 1$, we get $2 = 5A \Rightarrow A = \dfrac{2}{5}$

Substituting $x = -4$, we get $-8 = 5B \Rightarrow B = -\dfrac{8}{5}$

Therefore,

$I = \int \left( \dfrac{2/5}{1 - x} - \dfrac{8/5}{4 + x} \right) dx$

$I = \dfrac{2}{5} \left( \dfrac{\log|1 - x|}{-1} \right) - \dfrac{8}{5} \log|4 + x| + c$

$I = -\dfrac{8}{5} \log|4 + x| - \dfrac{2}{5} \log|1 - x| + c$

Answer: $-\dfrac{8}{5}\log|4 + x| - \dfrac{2}{5}\log|1 - x| + c$

Hence, the correct answer is option: 'D'.

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