find-1-dx-x-2-1-x-x-2-

Question Number 64866 by mathmax by abdo last updated on 22/Jul/19

f i n d \int_{1}^{+ \infty} \frac{d x}{x^{2} \sqrt{1 + x + x^{2}}}

Commented by mathmax by abdo last updated on 23/Jul/19

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

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

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

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

= 8 \int_{l n (2 + \sqrt{3})}^{+ \infty} \frac{d t}{3 c h (2 t) - 3 - 4 \sqrt{3} s h (t) + 2}

= 8 \int_{l n (2 + \sqrt{3})}^{+ \infty} \frac{d t}{3 c h (2 t) - 4 \sqrt{3} s h (t) - 1}

= 8 \int_{l n (2 + \sqrt{3})}^{+ \infty} \frac{d t}{3 \frac{e^{2 t} + e^{- 2 t}}{2} - 4 \sqrt{3} \frac{e^{t} - e^{- t}}{2} - 1}

= 16 \int_{l n (2 + \sqrt{3})}^{+ \infty} \frac{d t}{3 e^{2 t} + 3 e^{- 2 t} - 4 \sqrt{3} e^{t} + 4 \sqrt{3} e^{- t} - 2}

=_{e^{t} = u} 16 \int_{2 + \sqrt{3}}^{+ \infty} \frac{d u}{u (3 u^{2} + 3 u^{- 2} - 4 \sqrt{3} u + 4 \sqrt{3} u^{- 1} - 2)}

= 16 \int_{2 + \sqrt{3}}^{+ \infty} \frac{d u}{3 u^{3} + 3 u^{- 1} - 4 \sqrt{3} u^{2} + 4 \sqrt{3} - 2 u}

= 16 \int_{2 + \sqrt{3}}^{+ \infty} \frac{u d u}{3 u^{4} + 3 - 4 \sqrt{3} u^{3} + 4 \sqrt{3} u - 2 u^{2}} l e t d e c o m p o s e

F (u) = \frac{u}{3 u^{4} - 4 \sqrt{3} u^{3} - 2 u^{2} + 4 \sqrt{3} u + 3} \dots . b e c o n t i n u e d \dots .

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

l e t u s c h a n g e u = \frac{1}{x} d u = - \frac{d x}{x^{2}}

t h e n w e h a v e I = \int_{0}^{1} \frac{u d u}{\sqrt{1 + u + u^{2}}} = \frac{1}{2} [\int_{0}^{1} \frac{2 u + 1}{\sqrt{1 + u + u^{2}}} d u - \int_{0}^{1} \frac{d u}{\sqrt{1 + u + u^{2}}}]

f o r t h e 2^{n d} i n t e g r a l J w e k n o w s t h a t 1 + u + u^{2} = \frac{3}{4} [{(\frac{2 u + 1}{\sqrt{3}})}^{2} + 1]

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

s o

J = \int_{0}^{1} \frac{d u}{\sqrt{1 + u + u^{2}}} = \int_{\frac{π}{6}}^{\frac{π}{3}} \sqrt{{t a n}^{2} v + 1} d v = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{d v}{c o s v} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{c o s v d v}{(1 - s i n v) (1 + s i n v)} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{2} . [\frac{c o s v}{1 - s i n v} + \frac{c o s v}{1 + s i n v}] d v

= {[\frac{1}{2} l n ∣ \frac{1 + s i n v}{1 - s i n v} ∣]}_{\frac{π}{6}}^{\frac{π}{3}} = l n (\frac{4 + 2 \sqrt{3}}{3})

N o w

I = \frac{1}{2} . ({[2 \sqrt{1 + u + u^{2}}]}_{0}^{1} - J) = \frac{1}{2} . [(2 \sqrt{3} - 2) - l n (\frac{4 + 2 \sqrt{3}}{3})]

Commented by mathmax by abdo last updated on 28/Jul/19

t h a n k y o u s i r .