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JEE Main 2019 · 11 Jan · Shift II

JEE Main 2019 Mathematics question on Indefinite Integration

If ∫x+12x−1dx=f(x)2x−1+C\int \dfrac{x+1}{\sqrt{2x-1}} dx = f(x)\sqrt{2x-1} + C, where CC is a constant of integration, then f(x)f(x) is equal to
  1. A.
    23(x+2)\dfrac{2}{3}(x+2)
  2. B.
    13(x+4)\dfrac{1}{3}(x+4)
  3. C.
    13(x−4)\dfrac{1}{3}(x-4)
  4. D.
    13(x+1)\dfrac{1}{3}(x+1)
Show answer and solution

Correct option: B

Solution

Substitute t = 2x - 1, so x = (t+1)/2 and dx = dt/2. The integral becomes ∫((t+1)/2 + 1)/√t · (dt/2) = (1/4)∫(t+3)/√t dt = (1/4)∫(t^(1/2) + 3t^(-1/2)) dt = (1/4)[(2/3)t^(3/2) + 6t^(1/2)] + C = (1/4)·(2t^(1/2))[(t/3) + 3] + C = (1/2)√t · (t+9)/3 + C = ((t+9)/6)√t + C.

Substituting back t = 2x-1: result = ((2x-1+9)/6)√(2x-1) + C = ((2x+8)/6)√(2x-1) + C = ((x+4)/3)√(2x-1) + C.

So f(x) = (1/3)(x+4).

[JEE Main 2019 · 11 Jan · Shift II]

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