The integral ∫ dx ( x + 4 ) ^ (8)/(7) ( x - 3 ) ^ (6)/(7) is equal to: (where C is a constant of integration)

Correct answer: A. ((x - 3)/(x + 4))(1)/(7) + C ⇒ ∫ ((x - 3)/(x + 4))(- 6)/(7) 1 (x + 4) ^ 2 dx . . . (i) Let (x - 3)/(x + 4) = t ^ 7 , 7 (x + 4) ^ 2 dx = 7 t ^ 6 dt . . . (ii) From (i) from (ii) , ∫ t ^ - 6 t ^ 6 dt = t + C

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Indefinite Integration

The integral ∫ dx ( x + 4 ) ^ (8)/(7) ( x - 3 ) ^ (6)/(7) is equal...

The integral

∫ \dfrac{dx}{{( x + 4 )}^{\dfrac{8}{7}}{( x - 3 )}^{\dfrac{6}{7}}}

is equal to: (where

C

is a constant of integration)

Options

A $\left(\dfrac{x - 3}{x + 4}\right)\dfrac{1}{7} + C$
B $\left(\dfrac{x - 3}{x + 4}\right)\dfrac{- 1}{7} + C$
C $\dfrac{1}{2}\left(\dfrac{x - 3}{x + 4}\right)\dfrac{3}{7} + C$
D $- \dfrac{1}{13}\left(\dfrac{x - 3}{x + 4}\right)\dfrac{- 13}{7} + C$

Correct answer

A. $\left(\dfrac{x - 3}{x + 4}\right)\dfrac{1}{7} + C$

Step-by-step solution

$⇒ ∫ \left(\dfrac{x - 3}{x + 4}\right)\dfrac{- 6}{7}\dfrac{1}{{\left(x + 4\right)}^{2}}$ dx . . . $\left(i\right)$

Let $\dfrac{x - 3}{x + 4} = {t}^{7}$ ,

$\dfrac{7}{{\left(x + 4\right)}^{2}}dx = 7{t}^{6}dt . . . \left(ii\right)$

From \left(i\right) $from \left(ii\right)$ ,

$∫ {t}^{- 6}{t}^{6}dt = t + C$

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