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JEE Main202125 Feb 2021Evening ShiftMathematicsIndefinite IntegrationActual

The integral ∫ e 3 log e 2 x + 5 e 2 log e 2 x e 4 log e x + 5 e 3 log e x - 7 e 2 log e x d x , x > 0 , is equal to (where c is a constant of integration)

Options

  1. Alog e x 2 + 5 x - 7 + c
  2. B4 log e x 2 + 5 x - 7 + c
  3. C1 4 log e x 2 + 5 x - 7 + c
  4. Dlog e x 2 + 5 x - 7 + c

Correct answer

B. 4 log e x 2 + 5 x - 7 + c

Step-by-step solution

∫ e 3 log e 2 x + 5 e 2 log e 2 x e 4 log e x + 5 e 3 log e x - 7 e 2 log e x d x , x > 0 = ∫ 2 x 3 + 5 2 x 2 x 4 + 5 x 3 - 7 x 2 d x = ∫ 4 x 2 2 x + 5 x 2 x 2 + 5 x - 7 d x = 4 ∫ d x 2 + 5 x - 7 x 2 + 5 x - 7 = 4 log e x 2 + 5 x - 7 + c

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