∫ (dx/( (((x−1)^3 (x+2)^5 ))^(1/4) )) ?


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Question Number 122838 by bemath last updated on 20/Nov/20

$\int \frac{d x}{\sqrt[4]{{(x - 1)}^{3} {(x + 2)}^{5}}} ?$
	Answered by liberty last updated on 20/Nov/20
	$because (((x−1)^3 (x+2)^5 ))^(1/4) = (x−1)(x+2) (((x+2)/(x−1)))^(1/4) let u = (((x+2)/(x−1)))^(1/4) ⇒ x = ((u^4 +2)/(u^4 −1)) { ((x−1=(3/(u^4 −1)))),((x+2=((3u^4 )/(u^4 −1)))) :} and dx = ((−12u^3 )/((u^4 −1)^2 )) du L(x)=−∫ (((u^4 −1)^2 .12u^3 )/(3.3u^4 (u^4 −1)^2 u)) du L(x)=−(4/3)∫ (du/u^2 ) = (4/(3u)) + c L(x)= (4/3)(((x−1)/(x+2)))^(1/4) + c.$
	$b e c a u s e \sqrt[4]{{(x - 1)}^{3} {(x + 2)}^{5}} = (x - 1) (x + 2) \sqrt[4]{\frac{x + 2}{x - 1}}$ $l e t u = \sqrt[4]{\frac{x + 2}{x - 1}} \Rightarrow x = \frac{u^{4} + 2}{u^{4} - 1}$ ${\begin{cases} x - 1 = \frac{3}{u^{4} - 1} \\ x + 2 = \frac{3 u^{4}}{u^{4} - 1} \end{cases} a n d d x = \frac{- 12 u^{3}}{{(u^{4} - 1)}^{2}} d u$ $L (x) = - \int \frac{{(u^{4} - 1)}^{2} . 12 u^{3}}{3.3 u^{4} {(u^{4} - 1)}^{2} u} d u$ $L (x) = - \frac{4}{3} \int \frac{d u}{u^{2}} = \frac{4}{3 u} + c$ $L (x) = \frac{4}{3} \sqrt[4]{\frac{x - 1}{x + 2}} + c .$
	Commented by bemath last updated on 20/Nov/20

	$g r e a t$
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