calculate-0-1-x-1-3-1-x-1-x-2-3-dx-

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

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})