The integral ∫dx(x+4)87(x-3)67 is equal to: (where C is a constant of integration)

Let I=∫x-3x+4-671(x+4)2dx Put x-3x+4=t7⇒7(x+4)2dx=7t6dt So, I=∫t-6t6dt=t+c=x-3x+417+c

The integral ∫dx(x+4)87(x-3)67 is equal to: (where C is a constant of integration)

Let I=∫x-3x+4-671(x+4)2dx Put x-3x+4=t7⇒7(x+4)2dx=7t6dt So, I=∫t-6t6dt=t+c=x-3x+417+c

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

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

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

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

$\frac{1}{2} {(\frac{x - 3}{x + 4})}^{3 / 7} + C$

answer is B.

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Detailed Solution

$\begin{array}{l} Let I = \int {(\frac{x - 3}{x + 4})}^{\frac{- 6}{7}} \frac{1}{(x + 4)^{2}} d x \\ Put \frac{x - 3}{x + 4} = t^{7} \Rightarrow \frac{7}{(x + 4)^{2}} d x = 7 t^{6} d t \\ So, I = \int t^{- 6} t^{6} d t = t + c = {(\frac{x - 3}{x + 4})}^{\frac{1}{7}} + c \end{array}$