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

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Question: Answered & Verified by Expert

The integral

\int \frac{d x}{{(x + 4)}^{\frac{8}{7}} {(x - 3)}^{\frac{6}{7}}}

is equal to: (where

C

is a constant of integration)

MathematicsIndefinite IntegrationJEE MainJEE Main 2020 (09 Jan Shift 1)

Options:

A ${(\frac{x - 3}{x + 4})}^{\frac{1}{7}} + C$
B ${(\frac{x - 3}{x + 4})}^{\frac{- 1}{7}} + C$
C $\frac{1}{2} {(\frac{x - 3}{x + 4})}^{\frac{3}{7}} + C$
D $- \frac{1}{13} {(\frac{x - 3}{x + 4})}^{\frac{- 13}{7}} + C$

Solution:

1435 Upvotes Verified Answer

The correct answer is:

{(\frac{x - 3}{x + 4})}^{\frac{1}{7}} + C

$\Rightarrow \int {(\frac{x - 3}{x + 4})}^{\frac{- 6}{7}} \frac{1}{{(x + 4)}^{2}} d x . . . (i)$

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

$\frac{7}{{(x + 4)}^{2}} d x = 7 t^{6} d t . . . (ii)$

From $(i)$ from $(ii)$ ,

$\int t^{- 6} t^{6} d t = t + C$

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