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2025 PEA Q19

∫12log⁡x dx\displaystyle \int_1^2 \log x\, dx.

Reveal answer and solution

Answer

B

Solution

  1. 1

    Integration by parts with u=log⁡xu = \log x, dv=dxdv = dx:

  2. 2
    ∫log⁡x dx=xlog⁡x−x+C. \int \log x\, dx = x\log x - x + C.
  3. 3

    Therefore,

  4. 4
    ∫12log⁡x dx=(2log⁡2−2)−(1⋅0−1)=2log⁡2−1=log⁡4−1. \int_1^2 \log x\, dx = (2\log 2 - 2) - (1\cdot 0 - 1) = 2\log 2 - 1 = \log 4 - 1.

Answer structure / marking notes

2log⁡2=log⁡42 \log 2 = \log 4 --- this rewriting is the only subtlety.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.