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NDA 2020 Maths · Mathematics · Integrals · hard

Using integration by parts twice, the definite integral ∫1^e(ln x)^2 dx evaluates to

Single correct option

Using integration by parts twice, the definite integral ∫1e(ln⁡x)2 dx\displaystyle\int_1^e(\ln x)^2\,dx evaluates to
  1. A. e−1e-1
  2. B. 2−e2-e
  3. C. e−2e-2
  4. D. e+2e+2
Show answer and solution

Answer: (C) e−2e-2

By parts, ∫(ln⁡x)2dx=x(ln⁡x)2−2∫ln⁡x dx=x(ln⁡x)2−2(xln⁡x−x)=x(ln⁡x)2−2xln⁡x+2x\int(\ln x)^2dx=x(\ln x)^2-2\int\ln x\,dx=x(\ln x)^2-2(x\ln x-x)=x(\ln x)^2-2x\ln x+2x. At x=ex=e: e−2e+2e=ee-2e+2e=e. At x=1x=1: 22. The difference is e−2e-2, which is positive since e≈2.72e\approx2.72. The option 2−e2-e reverses the limits, and e−1e-1 is the value of ∫1eln⁡x dx\int_1^e\ln x\,dx only.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake