The integral ∫ 1 ( x - 1 ) ^ 3 ( x + 2 ) ^ 5 4 dx is equal to :...

$I = ∫ \dfrac{dx}{( x - 1{)}^{3} ( x + 2{)}^{5}4}$

$I = ∫ \dfrac{dx}{( x - 1{)}^{\dfrac{3}{4}} ( x + 2{)}^{\dfrac{5}{4}}}$

$I = ∫ \dfrac{dx}{( x - 1{)}^{2}{\left(\dfrac{x + 2}{x - 1}\right)}^{\dfrac{5}{4}}}$

Let $\dfrac{x + 2}{x - 1} = t$

Differentiate on both sides

$\dfrac{\left(x - 1\right) - \left(x + 2\right)}{{\left(x - 1\right)}^{2}} . dx = dt$

$⇒ \dfrac{dx}{{\left(x - 1\right)}^{2}} = \dfrac{- dt}{3}$

$I = \dfrac{- 1}{3} ∫ \dfrac{dt}{{\left(t\right)}^{\dfrac{5}{4}}}$

$= \dfrac{- 1}{3}\dfrac{{t}^{\dfrac{- 5}{4} + 1}}{1 - \dfrac{5}{4}} + C$

$= \dfrac{4}{3}t\dfrac{- 1}{4} + C$

$= \dfrac{4}{3}\left(\dfrac{1}{t}\right)\dfrac{1}{4} + C$

$= \dfrac{4}{3}\left(\dfrac{x - 1}{x + 2}\right)\dfrac{1}{4} + C$