Verified Solution • 2025–26 Session moderate

QJEEAdv06Ma8eca: JEE Advanced 2006 Mathematics - Class 12 JEEAdv06Ma8eca Solution: ∫ x 2 − 1 x 3 2 x 4 − 2 x 2 + 1 d x =

∫x2−1x32x4−2x2+1dx=

Multiple Choice Options

  • A 2x4−2x2+1x2+c
  • B 2x4−2x2+1x3+c
  • C 2x4−2x2+1x+c
  • D 2x4−2x2+12x2+c

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