calculate ∫_0 ^∞ (((1+x)^(1/3) −1)/(x(1+x)^(2/3) ))dx


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Question Number 43904 by abdo.msup.com last updated on 17/Sep/18

$c a l c u l a t e \int_{0}^{\infty} \frac{{(1 + x)}^{\frac{1}{3}} - 1}{x {(1 + x)}^{\frac{2}{3}}} d x$
Commented by maxmathsup by imad last updated on 19/Sep/18
$changement (1+x)^(1/3) =t give 1+x =t^3 ⇒ I =∫_1 ^(+∞) ((t−1)/((t^3 −1)t^2 )) (3t^2 )dt = ∫_1 ^(+∞) (3/(t^2 +t +1))dt = ∫_1 ^(+∞) ((3dt)/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3) ∫_(√3) ^(+∞) (3/((1+u^2 )))((√3)/2)du =2(√3) ∫_(√3) ^(+∞) (du/(1+u^2 )) =2(√3)[arctan(u)]_(√3) ^(+∞) =2(√3){(π/2)−(π/3)}=2(√3).(π/6) I = ((π(√3))/3) .$
$c h a n g e m e n t {(1 + x)}^{\frac{1}{3}} = t g i v e 1 + x = t^{3} \Rightarrow$ $I = \int_{1}^{+ \infty} \frac{t - 1}{(t^{3} - 1) t^{2}} (3 t^{2}) d t = \int_{1}^{+ \infty} \frac{3}{t^{2} + t + 1} d t$ $= \int_{1}^{+ \infty} \frac{3 d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} =_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{\sqrt{3}}^{+ \infty} \frac{3}{(1 + u^{2})} \frac{\sqrt{3}}{2} d u$ $= 2 \sqrt{3} \int_{\sqrt{3}}^{+ \infty} \frac{d u}{1 + u^{2}} = 2 \sqrt{3} {[a r c t a n (u)]}_{\sqrt{3}}^{+ \infty} = 2 \sqrt{3} {\frac{π}{2} - \frac{π}{3}} = 2 \sqrt{3} . \frac{π}{6}$ $I = \frac{π \sqrt{3}}{3} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18

	$1 + x = t^{3} d x = 3 t^{2} d t$ $\int_{1}^{\infty} \frac{t - 1}{(t^{3} - 1) (t^{2})} d t$ $\int_{1}^{\infty} \frac{d t}{t^{2} (t^{2} + t + 1)}$ $\frac{1}{t^{2} (t^{2} + t + 1)} = \frac{a}{t} + \frac{b}{t^{2}} + \frac{c t + d}{t^{2} + t + 1}$ $1 = a t (t^{2} + t + 1) + b (t^{2} + t + 1) + (c t + d) t^{2}$ $1 = a (t^{3} + t^{2} + t) + b (t^{2} + t + 1) + ({c t}^{3} + {d t}^{2})$ $1 = t^{3} (a + c) + t^{2} (a + b + d) + t (a + b) + b$ $b = 1$ $a + c = 0$ $a + 1 + d = 0$ $a + 1 = 0 a = - 1$ $c = 1$ $- 1 + 1 + d = 0 d = 0$ $a = - 1 b = 1 c = 1 d = 0$ $\int_{1}^{\infty} \frac{d t}{t^{2} (t^{2} + t + 1)} = \int_{1}^{\infty} \frac{- 1}{t} d t + \int_{1}^{\infty} \frac{1}{t^{2}} d t + \int_{1}^{\infty} \frac{t}{t^{2} + t + 1} d t$ $=∣ - l n t ∣_{1}^{\infty} + ∣ (\frac{- 1}{t}) ∣_{1}^{\infty} + \frac{1}{2} \int_{1}^{\infty} \frac{2 t + 1 - 1}{t^{2} + t + 1} d t$ $= (- l n \infty) + 1 + \frac{1}{2} \int_{1}^{\infty} \frac{d (t^{2} + t + 1)}{t^{2} + t + 1} - \frac{1}{2} \int_{1}^{\infty} \frac{d t}{t^{2} + 2 . t . \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}}$ $= (- l n \infty) + 1 + \frac{1}{2} ∣ l n (t^{2} + t + 1) ∣_{1}^{\infty} - \frac{1}{2} \int_{1}^{\infty} \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3} [}{2})}^{2}}$ $= (- l n \infty) + 1 + \frac{1}{2} l n (\infty) + \frac{1}{2} l n 3 - \frac{1}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} ∣ {t a n}^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) ∣_{1}^{\infty}$ $= (- \frac{1}{2} l n \infty) + 1 + \frac{1}{2} l n 3 - \frac{1}{\sqrt{3}} (\frac{Π}{2} - \frac{Π}{3})$ $= (- \frac{1}{2} l n \infty) + 1 + \frac{1}{2} l n 3 - \frac{1}{\sqrt{3}} (\frac{Π}{6})$
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