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

Question Number 43905 by abdo.msup.com last updated on 17/Sep/18

c a l c u l a t e \int_{1}^{+ \infty} \frac{d x}{x^{3} \sqrt{2 + x^{2}}}

Commented by maxmathsup by imad last updated on 18/Sep/18

l e t A = \int_{1}^{+ \infty} \frac{d x}{x^{3} \sqrt{2 + x^{2}}} d x c h a n g e m e n t x = \sqrt{2} s h (t) g i v e

A = \int_{a r g s h (\frac{1}{\sqrt{2}})}^{+ \infty} \frac{\sqrt{2} c h (t) d t}{2 \sqrt{2} {s h}^{3} (t) \sqrt{2} c h (t)} = \frac{1}{2 \sqrt{2}} \int_{l n (\frac{1}{\sqrt{2}} + \sqrt{\frac{3}{2}})}^{+ \infty} \frac{d t}{{(\frac{e^{t} - e^{- t}}{2})}^{3}}

= \frac{4}{\sqrt{2}} \int_{l n (\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}})}^{+ \infty} \frac{d t}{e^{3 t} + 3 e^{2 t} e^{- t} + 3 e^{t} e^{- 2 t} + e^{- 3 t}}

=_{e^{t} = u} 2 \sqrt{2} \int_{\frac{1 + \sqrt{3}}{\sqrt{2}}}^{+ \infty} \frac{1}{u^{3} + 3 u + 3 u^{- 1} + u^{- 3}} \frac{d u}{u}

= 2 \sqrt{2} \int_{\frac{1 + \sqrt{3}}{\sqrt{2}}}^{+ \infty} \frac{d u}{u^{4} + 3 u^{2} + 3 + u^{- 2}} = 2 \sqrt{2} \int_{\frac{1 + \sqrt{3}}{\sqrt{2}}}^{+ \infty} \frac{u^{2}}{u^{6} + 3 u^{4} + 3 u^{2} + 1} d u

l e t d e c o m p o s e F (u) = \frac{u^{2}}{u^{6} + 3 u^{4} + 3 u^{2} + 1} w e h a v e F (u) = \frac{u^{2}}{{(u^{2} + 1)}^{3}}

F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{{(u^{2} + 1)}^{2}} + \frac{e u + f}{{(u^{2} + 1)}^{3}}

F (- u) = F (u) \Rightarrow a = c = e = 0 \Rightarrow F (u) = \frac{b}{u^{2} + 1} + \frac{d}{{(u^{2} + 1)}^{2}} + \frac{f}{{(u^{2} + 1)}^{3}}

{l i m}_{u \to + \infty} u^{2} F (u) = 0 = b \Rightarrow F (u) = \frac{d}{{(u^{2} + 1)}^{2}} + \frac{f}{{(u^{2} + 1)}^{3}}

{l i m}_{u \to + \infty} u^{4} F (u) = 1 = d \Rightarrow F (u) = \frac{1}{{(u^{2} + 1)}^{2}} + \frac{f}{{(u^{2} + 1)}^{3}}

F (0) = 0 = 1 + f \Rightarrow f = - 1 \Rightarrow F (u) = \frac{1}{{(u^{2} + 1)}^{2}} - \frac{1}{{(u^{2} + 1)}^{3}} \Rightarrow

\int F (u) d u = \int \frac{d u}{{(u^{2} + 1)}^{2}} - \int \frac{d u}{{(u^{2} + 1)}^{3}} c h a n g e m e n t u = t a n θ g i v e

\int \frac{d u}{{(1 + u^{2})}^{2}} = \int \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{2}} d θ = \int {c o s}^{2} θ d θ = \int \frac{1 + c o s (2 θ)}{2} d θ

= \frac{θ}{2} + \frac{1}{4} s i n (2 θ) = \frac{θ}{2} + \frac{1}{4} \frac{2 t a n θ}{1 + {t a n}^{2} θ} = \frac{a r c t a n (u)}{2} + \frac{u}{1 + u^{2}}

= \frac{a r c t a n (e^{t})}{2} + \frac{e^{t}}{1 + e^{2 t}} = \dots . b e c o n t i n u e d \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18

x = \sqrt{2} t a n α d x = \sqrt{2} {s e c}^{2} α d α

\int_{{t a n}^{- 1} (\frac{1}{\sqrt{2}})}^{\frac{Π}{2}} \frac{\sqrt{2} {s e c}^{2} α d α}{2^{\frac{3}{2}} {t a n}^{3} α \times \sqrt{2 + 2 {t a n}^{2} α}}

\int_{{t a n}^{- 1} (\frac{1}{\sqrt{2}})}^{\frac{Π}{2}} \frac{\sqrt{2} {s e c}^{2} α}{2^{2} \times {t a n}^{3} α \times s e c α} d α

\frac{1}{2 \sqrt{2}} \int_{{t a n}^{- 1} (\frac{1}{\sqrt{2}})}^{\frac{Π}{2}} \frac{s e c α}{{t a n}^{3} α} d α

\frac{1}{2 \sqrt{2}} \int_{{t a n}^{- 1} (\frac{1}{\sqrt{2_{}}})}^{\frac{Π}{2}} \frac{{c o s}^{2} α}{{s i n}^{3} α} d α

o r a p p r o a c h

\int_{1}^{\infty} \frac{d x}{x^{3} \sqrt{2 + x^{2}}}

\int_{1}^{\infty} \frac{x d x}{x^{4} \sqrt{2 + x^{2}}}

t^{2} = 2 + x^{2} 2 t d t = 2 x d x

\int \frac{t d t}{{(t^{2} - 2)}^{2} \times t}

\int \frac{d t}{{(t + \sqrt{2})}^{2} \times {(t - \sqrt{2})}^{2}}

\frac{1}{{(t^{2} - 2)}^{2}} = \frac{a}{t + \sqrt{2}} + \frac{b}{{(t + \sqrt{2})}^{2}} + \frac{c}{t - \sqrt{2}} + \frac{d}{{(t - \sqrt{2})}^{2}}

1 = a (t + \sqrt{2}) {(t - \sqrt{2})}^{2} + b {(t - \sqrt{2})}^{2} + c (t - \sqrt{2}) {(t + \sqrt{2})}^{2} + d {(t + \sqrt{2})}^{2}

t + \sqrt{2} = 0 1 = b {(- 2 \sqrt{2})}^{2} b = \frac{1}{8}

t - \sqrt{2} = 0 1 = d {(2 \sqrt{2})}^{2} d = \frac{1}{8}

t = 0 1 = a \times 2 \sqrt{2} + \frac{1}{8} \times 2 + c (- 2 \sqrt{2}) + \frac{1}{8} \times 2

1 - \frac{1}{2} = 2 \sqrt{2} (a - c)

a - c = \frac{1}{4 \sqrt{2}}

t = 2 \sqrt{2}

1 = a (3 \sqrt{2}) (2) + \frac{1}{8} (2) + c (\sqrt{2}) (18) + \frac{1}{8} \times 18

1 - \frac{1}{4} - \frac{9}{4} = 6 \sqrt{2} a + 18 \sqrt{2} c

\frac{4 - 1 - 9}{4} = 6 \sqrt{2} (a + 3 c)

a + 3 c = \frac{- 6}{4 \times 6 \sqrt{2}} = \frac{- 1}{4 \sqrt{2}}

a - c = \frac{1}{4 \sqrt{2}}

a + 3 c = \frac{- 1}{4 \sqrt{2}}

- 4 c = \frac{2}{4 \sqrt{2}}

c = \frac{- 1}{8 \sqrt{2}}

a = \frac{1}{4 \sqrt{2}} - \frac{1}{8 \sqrt{2}} = \frac{1}{8 \sqrt{2}}

\int \frac{a}{t + \sqrt{2}} d t + \int \frac{b}{{(t + \sqrt{2})}^{2}} d t + \int \frac{c}{t - \sqrt{2}} d t + \int \frac{d}{{(t - \sqrt{2})}^{2}}

\frac{1}{8 \sqrt{2}} l n (t + \sqrt{2}) + \frac{1}{8} \times \frac{- 1}{(t + \sqrt{2})} + \frac{- 1}{8 \sqrt{2}} l n (t - \sqrt{2}) + \frac{1}{8} \times \frac{- 1}{(t - \sqrt{2})}

\frac{1}{8 \sqrt{2}} l n (\frac{t + \sqrt{2}}{t - \sqrt{2}}) - \frac{1}{8} (\frac{1}{t + \sqrt{2}} + \frac{1}{t - \sqrt{2}})

\frac{1}{8 \sqrt{2}} l n (\frac{\sqrt{2 + x^{2}} + \sqrt{2}}{\sqrt{2 + x^{2}} - \sqrt{2}}) - \frac{1}{8} (\frac{2 \sqrt{2 + x^{2}}}{2 + x^{2} - 2})

a n s i s

∣ \frac{1}{8 \sqrt{2}} l n (\frac{\sqrt{2 + x^{2}} + \sqrt{2}}{\sqrt{2 + x^{2}} - \sqrt{2}}) - \frac{1}{4} (\frac{\sqrt{2 + x^{2}}}{x^{2}}) ∣_{1}^{\infty}

= \frac{1}{8 \sqrt{2}} ∣ l n (\frac{1 + \frac{\sqrt{2}}{\sqrt{2 + x^{2}}}}{1 - \frac{\sqrt{2}}{\sqrt{2 + x^{2}}}}) - \frac{1}{4} (\frac{x}{x^{2}} \sqrt{1 + \frac{2}{x^{2}}}) ∣_{1}^{\infty}

= \frac{1}{8 \sqrt{2}} [{l n (\frac{1 + 0}{1 - 0}) - \frac{1}{4} (0 \times \sqrt{1 + 0}} - {l n (\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}) - \frac{1}{4} (\sqrt{3})

\frac{1}{8 \sqrt{2}} [\frac{\sqrt{3}}{4} - l n (\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}})]