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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ x ^ pq - p - 1 ( x ^ p + 1) ^ q dx = 2 ( 1+ x ^ - p ) ^ 1 - q...

If ∫xpq−p−1(xp+1)q∫ \dfrac{{x}^{pq - p - 1}}{{\left({x}^{p} + 1\right)}^{q}}dx = \dfrac{2 ( 1+{x}^{- p}{)}^{1 - q}}{\lambdap ( q - 1 )} + c\left(p, q \in N - { 1 }\right),thenthevalueof, then the value of\lambdais(here,is (here,c is an arbitrary constant)

Correct answer

2

Step-by-step solution

I=∫x−p−1(1+x−p)qdxI = ∫ \dfrac{{x}^{- p - 1}}{( 1 + {x}^{- p}{)}^{q}}dx

Let, 1+x−p=t1 + {x}^{- p} = t

=−1p∫dttq= \dfrac{- 1}{p} ∫ \dfrac{dt}{{t}^{q}}

=t−q+1p(1−q)+c= \dfrac{{t}^{- q + 1}}{p ( 1 - q )} + c

∴2λ=1∴ \dfrac{2}{\lambda} = 1

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