∫dxx+4x−387×x−367×x−387=∫x+4x−3−87x−32 dxPut =x+4x−3=t−7x−32dx=dt=∫t−87−dt7=−17t−17−17+c=x+4x−3−17+c=x−3x+417+c

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

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

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

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

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

answer is C.

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Detailed Solution

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

$= \int \frac{{(\frac{x + 4}{x - 3})}^{\frac{- 8}{7}}}{{(x - 3)}^{2}} d x$

Put $= \frac{x + 4}{x - 3} = t$

$\frac{- 7}{{(x - 3)}^{2}} d x = d t$

$= \int t^{\frac{- 8}{7}} (\frac{- d t}{7}) = \frac{- 1}{7} (\frac{t^{\frac{- 1}{7}}}{\frac{- 1}{7}}) + c$

$= {(\frac{x + 4}{x - 3})}^{\frac{- 1}{7}} + c = {(\frac{x - 3}{x + 4})}^{\frac{1}{7}} + c$

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