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If ∫ x p q - p - 1 x p + 1 q d x = 2 ( 1+ x - p ) 1 - q λ p ( q - 1 ) + c    p , q ∈ N - 1 , then the value of λ is (here, c is an arbitrary constant)

Correct answer

2

Step-by-step solution

I = ∫ x - p - 1 ( 1 + x - p ) q d x Let, 1 + x - p = t = - 1 p ∫ d t t q = t - q + 1 p ( 1 - q ) + c ∴ 2 λ = 1

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