If 1 ( x - 1 ) 4 ( x + 3 ) 6 5 d x = A ( x - 1 x + 3 ) B + C , where C is the constant of integration, then the value of + 20 A B is ____ . [2024]

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Solution

Q.
$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

Ans.
(7)

$I = \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x$

Putting $\frac{x + 3}{x - 1} = t \Rightarrow x = (\frac{3 + t}{t - 1}) \Rightarrow d x = \frac{- 4}{{(t - 1)}^{2}} d t$

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

$∴ I = \int \frac{\frac{- 4}{{(t - 1)}^{2}} d t}{t^{1 / 5} (\frac{3 + t}{t - 1} - 1) (\frac{3 + t}{t - 1} + 3)}$

$I = \int \frac{- 4 d t}{t^{1 / 5} (16 t)} = \frac{- 1}{4} \int \frac{d t}{t^{6 / 5}} = \frac{- 1}{4} (\frac{t^{- 1 / 5}}{- 1 / 5}) + C$

On comparing, we get

$A = \frac{5}{4}, B = \frac{1}{5}, α = 1, β = 1$

Hence, $α + β + 20 A B = 1 + 1 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7$

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