NDA 2024 · PYQ · Integration · medium
Let ∫ dx/(√(x+1) - √(x-1)) = α(x+1)^(3/2) + β(x-1)^(3/2) + c. What is the value of α?
Answer
The correct answer is A: 1/3. Rationalize: 1/(√(x+1) - √(x-1)) × (√(x+1) + √(x-1))/(√(x+1) + √(x-1)) = (√(x+1) + √(x-1))/((x+1) - (x-1)) = (√(x+1) + √(x-1))/2. So integral = (1/2)∫[√(x+1) + √(x-1)] dx = (1/2)[(2/3)(x+1)^(3/2) + (2/3)(x-1)^(3/2)] + c = (1/3)(x+1)^(3/2) + (1/3)(x-1)^(3/2) + c.
- A.1/3✓ Correct
- B.2/3
- C.1
- D.4/3
Rationalize: 1/(√(x+1) - √(x-1)) × (√(x+1) + √(x-1))/(√(x+1) + √(x-1)) = (√(x+1) + √(x-1))/((x+1) - (x-1)) = (√(x+1) + √(x-1))/2. So integral = (1/2)∫[√(x+1) + √(x-1)] dx = (1/2)[(2/3)(x+1)^(3/2) + (2/3)(x-1)^(3/2)] + c = (1/3)(x+1)^(3/2) + (1/3)(x-1)^(3/2) + c. So α = 1/3.
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