∫ d x ( x − 3 ) ( 4 / 5 ) ( x + 1 ) 6 / 5 = Options: A . &nbsp ( ( x − 3 ) ( x + 1 ) ( 1 / 5 ) ) + c B . &nbsp 5 4 ( x − 3 x + 1 ) 1 / 5 + c C . &nbsp ( x − 3 x + 1 ) 1 / 5 + c D . &nbsp ( x − 3 ) 6 / 5 ( x + 1 ) 4 / 5 + c

Answer: Option B : B I = ∫ d x ( x − 3 ) 4 / 5 ( x + 1 ) 6 / 5 P u t t = x − 3 x + 1 d t = 4 ( x + 1 ) 2 d x . I = 1 4 ∫ d t t 4 / 5 = 1 4 ( t − 4 / 5 + 1 − 4 / 5 + 1 ) = 5 4 t 1 / 5 = 5 4 ( x − 3 x + 1 ) 1 / 5

∫dx(x−3)(4/5)(x+1)6/5=

Discussion Forum : Indefinite Integration

Question -

$\int \frac{d x}{(x - 3)^{(4 / 5)} (x + 1)^{6 / 5}} =$

Options:

A .

((x - 3) (x + 1)^{(1 / 5)}) + c

B .

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

C .

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

D .

(x - 3)^{6 / 5} (x + 1)^{4 / 5} + c

Answer: Option B
:
B

I = \int \frac{d x}{(x - 3)^{4 / 5} (x + 1)^{6 / 5}} P u t t = \frac{x - 3}{x + 1} d t = \frac{4}{(x + 1)^{2}} d x . I = \frac{1}{4} \int \frac{d t}{t^{4 / 5}} = \frac{1}{4} (\frac{t^{- 4 / 5} + 1}{- 4 / 5} + 1) = \frac{5}{4} t^{1 / 5} = \frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5}

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Discussion Forum : Indefinite Integration

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