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$\int \frac{d x}{{(x + 4)}^{8 / 7} {(x - 3)}^{6 / 7}}$ is equal to (where C is a constant of integration)

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

$- \frac{1}{13} {(\frac{x - 3}{x + 4})}^{1 / 7} + C$

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

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

answer is A.

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Detailed Solution

Let $\frac{x + 4}{x - 3} = t \Rightarrow \frac{d x}{{(x - 3)}^{2}} = \frac{1}{7} d t \Rightarrow I = - \frac{1}{7} \int t^{- 8 / 7} d t = t^{- 1 / 7} + C$

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