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JEE Main 2019 · 12 April · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

A value of α\alpha such that ∫αα+1dx(x+α)(x+α+1)=log⁡e(98)\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_e\left(\frac{9}{8}\right) is
  1. A.
    −2-2
  2. B.
    12\frac{1}{2}
  3. C.
    −12-\frac{1}{2}
  4. D.
    22
Show answer and solution

Correct option: A

Solution

Use partial fractions: 1/[(x+α)(x+α+1)] = 1/(x+α) - 1/(x+α+1).

Integrating from α to α+1: [ln(x+α) - ln(x+α+1)] from α to α+1

= [ln(2α+1) - ln(2α+2)] - [ln(2α) - ln(2α+1)]

= 2·ln(2α+1) - ln(2α) - ln(2α+2)

= ln[(2α+1)² / (2α(2α+2))]

Set equal to ln(9/8): (2α+1)² / [2α·(2α+2)] = 9/8.

For α = -2: (2(-2)+1)² / [2(-2)·(2(-2)+2)] = (-3)² / [(-4)(-2)] = 9/8. ✓

[JEE Main 2019 · 12 April · Shift II]

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