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

Question Number 41302 by math khazana by abdo last updated on 05/Aug/18

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

Commented by maxmathsup by imad last updated on 05/Aug/18

l e t I = \int_{1}^{+ \infty} \frac{d x}{x^{2} \sqrt{x^{2} + x + 1}}

I = \int_{1}^{+ \infty} \frac{d x}{x^{2} \sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ g i v e

I = \int_{a r c t a n (\frac{1}{\sqrt{3}})}^{\frac{π}{2}} \frac{1}{{(\frac{\sqrt{3}}{2} t a n θ - \frac{1}{2})}^{2} \frac{\sqrt{3}}{2}} {c o s}^{2} θ \frac{\sqrt{3}}{2} (1 + {t a n}^{2} θ) d θ

= \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{4 d θ}{{(\sqrt{3} t a n θ - 1)}^{2}} = \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{4}{3 {t a n}^{2} θ - 2 \sqrt{3} t a n θ + 1} d θ

=_{t a n θ = x} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{4}{(3 x^{2} - 2 \sqrt{3 x} + 1)} \frac{d x}{1 + x^{2}} l e t d e c o m p o s e

F (x) = \frac{4}{(x^{2} + 1) (3 x^{2} - 2 \sqrt{3} x + 1)} = \frac{4}{{(\sqrt{3} x - 1)}^{2} (x^{2} + 1)}

F (x) = \frac{a}{\sqrt{3} x - 1} + \frac{b}{{(\sqrt{3} x - 1)}^{2}} + \frac{b x + c}{x^{2} + 1}

b = {l i m}_{x \to \frac{1}{\sqrt{3}}} {(\sqrt{3} x - 1)}^{2} F (x) = \frac{4}{\frac{4}{3}} = 3

{l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{\sqrt{3}} + b \Rightarrow b = - \frac{a}{\sqrt{3}} \Rightarrow

F (x) = \frac{a}{\sqrt{3} x - 1} + \frac{3}{{(\sqrt{3} x - 1)}^{2}} + \frac{- \frac{a}{\sqrt{3}} x + c}{x^{2} + 1}

F (0) = - a + 3 + c \Rightarrow c = a - 3 \Rightarrow

F (x) = \frac{a}{\sqrt{3} x - 1} + \frac{3}{{(\sqrt{3} x - 1)}^{2}} + \frac{- \frac{a}{\sqrt{3}} x + a - 3}{x^{2} + 1}

F (\sqrt{3}) = \frac{4}{4 (4)} = \frac{1}{4} = \frac{a}{2} + \frac{3}{4} + \frac{- 3}{4} \Rightarrow \frac{a}{2} = \frac{1}{4} \Rightarrow a = \frac{1}{2} \Rightarrow

F (x) = \frac{1}{2 (\sqrt{3} x - 1)} + \frac{3}{{(\sqrt{3} x - 1)}^{2}} + \frac{- \frac{1}{2 \sqrt{3}} x - \frac{5}{2}}{x^{2} + 1} \Rightarrow

\int F (x) d x = \frac{1}{2 \sqrt{3}} \int \frac{d x}{x - \frac{1}{\sqrt{3}}} + \int \frac{d x}{{(x - \frac{1}{\sqrt{3}})}^{2}} - \frac{1}{2 \sqrt{3}} \int \frac{x + 5 \sqrt{3}}{x^{2} + 1} d x

= \frac{1}{2 \sqrt{3}} l n ∣ x - \frac{1}{\sqrt{3}} ∣ - \frac{1}{x - \frac{1}{\sqrt{3}}} - \frac{1}{4 \sqrt{3}} \int \frac{2 x}{x^{2} + 1} - \frac{5}{2} \int \frac{d x}{x^{2} + 1}

= \frac{1}{2 \sqrt{3}} l n ∣ x - \frac{1}{\sqrt{3}} ∣ - \frac{1}{x - \frac{1}{\sqrt{3}}} - \frac{1}{4 \sqrt{3}} l n (x^{2} + 1) - \frac{5}{2} a r c t a n x + c \Rightarrow

\int_{\frac{1}{\sqrt{3}}}^{+ \infty} F (x) d x = \dots . (p u t t h e l m i t s)

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

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

t = \frac{1}{x} d t = - \frac{1}{x^{2}} d x

\int_{1}^{0} \frac{- d t}{\sqrt{\frac{1}{t^{2}} (1 + t + t^{2})}}

\int_{0}^{1} \frac{t d t}{\sqrt{t^{2} + t + 1}}

= \frac{1}{2} \int_{0}^{1} \frac{2 t + 1 - 1}{\sqrt{t^{2} + t + 1}} d t

= \frac{1}{2} \int_{0}^{1} \frac{2 t + 1}{\sqrt{t^{2} + t + 1}} - \frac{1}{2} \int_{0}^{1} \frac{d t}{\sqrt{t^{2} + 2 . t . \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}}}

= \frac{1}{2} \int_{0}^{1} \frac{d (t^{2} + t + 1)}{\sqrt{t^{2} + t + 1}} - \frac{1}{2} \int_{0}^{1} \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}}

= \frac{1}{2} ∣ \frac{\sqrt{t^{2} + t + 1}}{\frac{1}{2}} ∣_{0}^{1} - \frac{1}{2} ∣ l n {(t + \frac{1}{2}) + \sqrt{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} ∣_{0}^{1}

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

= (\sqrt{3} - 1) - \frac{1}{2} {l n (\frac{3 + \sqrt{6}}{2}) - l n (\frac{3}{2})}

= (\sqrt{3} - 1) - \frac{1}{2} {l n (\frac{3 + \sqrt{6}}{3})}