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Q.

$\int \frac{d x}{{(x - 2)}^{7 / 8} {(x + 3)}^{9 / 8}} =$

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a

$\frac{8}{5} {(\frac{x - 2}{x + 3})}^{1 / 8} + c$

b

$\frac{8}{5} {(\frac{x + 2}{x + 3})}^{1 / 8} + c$

c

$\frac{4}{5} {(\frac{x - 2}{x + 3})}^{1 / 8} + c$

d

$\frac{5}{8} {(\frac{x - 2}{x + 3})}^{1 / 8} + c$

answer is A.

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Detailed Solution

$\int \frac{d x}{{(x - 2)}^{7 / 8} {(x + 3)}^{9 / 8}}$
$\int \frac{d x}{{(\frac{x - 2}{x + 3})}^{\frac{7}{8}} {(x + 3)}^{2}} p u t \frac{x - 2}{x + 3} = t \Rightarrow \frac{5}{{(x + 3)}^{2}} d x = d t \Rightarrow \frac{1}{5} \int t^{- \frac{7}{8}} d t = \frac{8}{5} t^{\frac{1}{8}} + c$
$= \frac{8}{5} {(\frac{x - 2}{x + 3})}^{\frac{1}{8}} + c$

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