If the integral I = ∫ e ^ 5ln x ( x ^ 6 + 1) ^ - 1 dx = ( x ^ 6 + 1) + C , (where, C is the constant of integration) then the value of (1)/(λ) is

Correct answer: 6 The integral I = ∫ e ^ 5ln x ( x ^ 6 + 1) ^ - 1 dx I = ∫ e ^ ( x ^ 5 ) x ^ 6 + 1 dx = ∫ x ^ 5 x ^ 6 + 1 dx = (1)/(6) ∫ 6 x ^ 5 dx x ^ 6 + 1 Put x ^ 6 + 1 = t 6 x ^ 5 dx = dt ⇒ I = (1)/(6) ∫ (dt)/(t) = (1)/(6) (t) + C ⇒ I = (1)/(6) ( x ^ 6 + 1) + C ⇒ λ = (1)/(6) ⇒ (1)/(λ) = 6

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral I = ∫ e ^ 5ln x ( x ^ 6 + 1) ^ - 1 dx = ( x ^ 6 +...

If the integral

I = ∫ {e}^{5ln x}{\left({x}^{6} + 1\right)}^{- 1}dx

= \ln \left({x}^{6} + 1\right) + C

, (where,

C

is the constant of integration) then the value of

\dfrac{1}{\lambda}

Correct answer

Step-by-step solution

The integral

I = ∫ {e}^{5ln x}{\left({x}^{6} + 1\right)}^{- 1}dx

I = ∫ \dfrac{{e}^{\ln \left({x}^{5}\right)}}{{x}^{6} + 1}dx

= ∫ \dfrac{{x}^{5}}{{x}^{6} + 1}dx

= \dfrac{1}{6} ∫ \dfrac{6{x}^{5}dx}{{x}^{6} + 1}

Put

{x}^{6} + 1 = t

6{x}^{5}dx = dt

⇒ I = \dfrac{1}{6} ∫ \dfrac{dt}{t} = \dfrac{1}{6}\ln \left(t\right) + C

⇒ I = \dfrac{1}{6}\ln \left({x}^{6} + 1\right) + C

⇒ \lambda = \dfrac{1}{6}

⇒ \dfrac{1}{\lambda} = 6

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