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Question Number 156422 by SANOGO last updated on 11/Oct/21

Commented by cortano last updated on 11/Oct/21
$let x=t^6 →dx=6t^5 dt I=∫ (t^3 /(1−t^2 )) (6t^5 dt) = ∫ ((6t^8 )/(1−t^2 )) dt partial fraction$
$let x = t^{6} \to dx = {6 t}^{5} dt$ $I = \int \frac{t^{3}}{1 - t^{2}} ({6 t}^{5} dt)$ $= \int \frac{{6 t}^{8}}{1 - t^{2}} dt$ $partial fraction$
Commented by SANOGO last updated on 11/Oct/21

$m e r c i b i e n$
	Answered by puissant last updated on 11/Oct/21
	$K=∫(1/( (√x)−(x)^(1/3) ))dx ; x=u^6 →dx=6u^5 du ⇒ K=∫((6u^5 )/(u^3 −u^2 ))du = 6∫(u^3 /(u−1))du =6∫((u^3 −1+1)/(u−1))du = 6∫((u^3 −1)/(u−1))du+6∫(1/(u−1))du =6∫{u^2 +u+1}du+6ln∣u∣ ⇒ K=2u^3 +3u^2 +6u+6ln∣u∣+C ⇒ K=2(x^(1/6) )^3 +3(x^(1/6) )^2 +6x^(1/6) +6ln∣x^(1/6) ∣+C ∴∵ K=2(√x)+3(x)^(1/3) +6(x)^(1/6) +ln∣x∣+C.. ..........Le puissant...........$
	$K = \int \frac{1}{\sqrt{x} - \sqrt[3]{x}} d x; x = u^{6} \to d x = 6 u^{5} d u$ $\Rightarrow K = \int \frac{6 u^{5}}{u^{3} - u^{2}} d u = 6 \int \frac{u^{3}}{u - 1} d u$ $= 6 \int \frac{u^{3} - 1 + 1}{u - 1} d u = 6 \int \frac{u^{3} - 1}{u - 1} d u + 6 \int \frac{1}{u - 1} d u$ $= 6 \int {u^{2} + u + 1} d u + 6 l n ∣ u ∣$ $\Rightarrow K = 2 u^{3} + 3 u^{2} + 6 u + 6 l n ∣ u ∣ + C$ $\Rightarrow K = 2 {(x^{\frac{1}{6}})}^{3} + 3 {(x^{\frac{1}{6}})}^{2} + 6 x^{\frac{1}{6}} + 6 l n ∣ x^{\frac{1}{6}} ∣ + C$ $∴∵ K = 2 \sqrt{x} + 3 \sqrt[3]{x} + 6 \sqrt[6]{x} + l n ∣ x ∣ + C . .$ $. . . . . . . . . . L e p u i s s a n t . . . . . . . . . . .$
	Commented by SANOGO last updated on 11/Oct/21

	$m e r c i b i e n$
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