Verified Solution • 2025–26 Session moderate

QJEEM18Ma708c: JEE Main 2018 Mathematics - Class 12 JEEM18Ma708c Solution: If (x R {1, 2}), then is equal to : (where C is a constant of…

If f(x−4x+2)=2x+1,f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1, (x ∈\in R −- {1, −- 2}), then ∫f(x)dx\int f \left( x \right)dx is equal to :
(where C is a constant of integration)

Multiple Choice Options

  • A 12 loge | 1 −- x | + 3x + C
  • B −- 12 loge | 1 −- x | −- 3x + C
  • C 12 loge | 1 −- x | −- 3x + C
  • D −- 12 loge | 1 −- x | + 3x + C

Step-by-Step Verified Solution

Let, x−4x+2{{{x - 4} \over {x + 2}}} = t

⇒\Rightarrow x - 4 = t(x+2)

⇒\Rightarrow x (1 -t) = 2(t+2)

⇒\Rightarrow x = 2(t+2)1−t{{2(t + 2)} \over {1 - t}}

∴\therefore f(t) = 2( 2(t+2)1−t{{2(t + 2)} \over {1 - t}} ) + 1

= 4t+81−t+1{{4t + 8} \over {1 - t}} + 1

= 3t+91−t{{3t + 9} \over {1 - t}}

= 3(t+3)1−t{{3(t + 3)} \over {1 - t}}

= 3(t−1+4)1−t{{3(t - 1 + 4)} \over {1 - t}}

= - 3 + 121−t12 \over 1 - t

∴\therefore f(x) = - 3 + 121−x12 \over 1 - x

∴\therefore ∫f(x)dx\int {f(x)dx}

= ∫(−3+121−x)dx\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx}

= −12log⁡e∣1−x∣-12{\log _e}|1 - x| - 3x + C

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