Evaluate the integral ∫ (((x)^(1/3) +1)/( (x)^(1/3) −1)) dx


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Question Number 122689 by john santu last updated on 19/Nov/20

$E v a l u a t e t h e i n t e g r a l$ $\int \frac{\sqrt[3]{x} + 1}{\sqrt[3]{x} - 1} d x$
	Answered by bobhans last updated on 19/Nov/20

	$l e t x = u^{3} \Rightarrow d x = 3 u^{2} d u$ $C (x) = \int \frac{u + 1}{u - 1} . (3 u^{2} d u)$ $C (x) = 3 \int \frac{u^{3} + u^{2}}{u - 1} d u$ $C (x) = 3 \int (u^{2} + 2 u + 2 + \frac{2}{u - 1}) d u$ $C (x) = 3 (\frac{1}{3} u^{3} + u^{2} + 2 u + 2 \ln ∣ u - 1 ∣) + c$ $C (x) = x + 3 \sqrt[3]{x^{2}} + 6 \sqrt[3]{x} + 6 \ln ∣ \sqrt[3]{x} - 1 ∣ + c$
	Commented by MJS_new last updated on 19/Nov/20

	$another possibility :$ $\int \frac{x^{1 / 3} + 1}{x^{1 / 3} - 1} d x = \int d x + \int \frac{2}{x^{1 / 3} - 1} d x$ $[t = x^{1 / 3} - 1 \to d x = 3 x^{2 / 3} d t]$ $= x + 6 \int \frac{{(t + 1)}^{2}}{t} d t = x + 3 t (t + 4) + 6 \ln t =$ $= x + 3 (x^{1 / 3} - 1) (x^{1 / 3} + 3) + 6 \ln ∣ x^{1 / 3} - 1 ∣ + C$
	Commented by john santu last updated on 19/Nov/20

	$t h a n k y o u$
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