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Question Number 34956 by behi83417@gmail.com last updated on 13/May/18

Commented by a.i msup by abdo last updated on 13/May/18
$changement (√x)=t give I = ∫_0 ^1 (t/(t^4 +t^2 +1)) 2tdt = 2∫_0 ^1 (t^2 /(t^4 +t^2 +1))dt ch. t=(1/u) I = −2∫_1 ^(+∞) (1/(u^2 ((1/u^4 ) +(1/u^2 ) +1)))(−(du/u^2 )) =2 ∫_1 ^(+∞) (du/(u^4 +u^2 +1)) 2I = 2{ ∫_0 ^1 ((t^2 )/(t^4 +t^2 +1))dt+∫_1 ^(+∞) (dt/(t^4 +t^2 +1))} ⇒ I = ∫_0 ^∞ ((1+t^2 )/(t^4 +t^2 +1))dt 2I = ∫_(−∞) ^(+∞) ((1+t^2 )/(t^4 +t^2 +1))dt let consider the complex function ϕ(z) = ((1+z^2 )/(z^4 +z^2 +1)) poles of ϕ?$
$c h a n g e m e n t \sqrt{x} = t g i v e$ $I = \int_{0}^{1} \frac{t}{t^{4} + t^{2} + 1} 2 t d t$ $= 2 \int_{0}^{1} \frac{t^{2}}{t^{4} + t^{2} + 1} d t c h . t = \frac{1}{u}$ $I = - 2 \int_{1}^{+ \infty} \frac{1}{u^{2} (\frac{1}{u^{4}} + \frac{1}{u^{2}} + 1)} (- \frac{d u}{u^{2}})$ $= 2 \int_{1}^{+ \infty} \frac{d u}{u^{4} + u^{2} + 1}$ $2 I = 2 {\int_{0}^{1} \frac{t^{2}}{t^{4} + t^{2} + 1} d t + \int_{1}^{+ \infty} \frac{d t}{t^{4} + t^{2} + 1}}$ $\Rightarrow I = \int_{0}^{\infty} \frac{1 + t^{2}}{t^{4} + t^{2} + 1} d t$ $2 I = \int_{- \infty}^{+ \infty} \frac{1 + t^{2}}{t^{4} + t^{2} + 1} d t$ $l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{1 + z^{2}}{z^{4} + z^{2} + 1} p o l e s o f φ ?$
Commented by abdo mathsup 649 cc last updated on 13/May/18
$let solve z^4 +z^2 +1 =0 changement z^2 =t ⇒ t^2 +t +1 =0 the roots t_1 =e^(i((2π)/3)) , t_2 = e^(−i((2π)/3)) ϕ(z) = ((1+z^2 )/((z^2 −e^(i((2π)/3)) )(z^2 −e^(−i((2π)/3)) ))) = ((1+z^2 )/((z −e^(i(π/3)) )(z +e^(i(π/3)) )(z −e^(−i(π/3)) )(z +e^(−i(π/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ, e^(i(π/3)) ) +Res(ϕ,−e^(−i(π/3)) )} Res(ϕ, e^(i(π/3)) )= ((1+ e^(i((2π)/3)) )/(2 e^(i(π/3)) (2i((√3)/2))(2(1/2)))) = (1/(2i(√3)))( e^(−i(π/3)) + e^(i(π/3)) ) = (1/(2i(√3))) 2.(1/2) =(1/(2i(√(3 )))) Res(ϕ ,−e^(−i(π/3)) ) = ((1+e^(−i((2π)/3)) )/((−e^(−i(π/3)) −e^(i(π/3)) )(−e^(−i(π/3)) +e^(i(π/3)) )(−2 e^(−i(π/3)) ))) = ((1+ e^(−i((2π)/3)) )/(2 ( 2(1/2))(2i((√3)/2))e^(−i(π/3)) )) = (1/(2i(√3))) ( e^(i(π/3)) +e^(−i(π/3)) ) = (1/(2i(√3))) (2(1/2)) = (1/(2i(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(2i(√3))) +(1/(2i(√3)))}=2iπ(1/(i(√3))) =((2π)/(√3)) 2I =((2π)/(√(3 ))) ⇒ I = (π/(√3)) .$
$l e t s o l v e z^{4} + z^{2} + 1 = 0 c h a n g e m e n t z^{2} = t \Rightarrow$ $t^{2} + t + 1 = 0 t h e r o o t s t_{1} = e^{i \frac{2 π}{3}}, t_{2} = e^{- i \frac{2 π}{3}}$ $φ (z) = \frac{1 + z^{2}}{(z^{2} - e^{i \frac{2 π}{3}}) (z^{2} - e^{- i \frac{2 π}{3}})}$ $= \frac{1 + z^{2}}{(z - e^{i \frac{π}{3}}) (z + e^{i \frac{π}{3}}) (z - e^{- i \frac{π}{3}}) (z + e^{- i \frac{π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{i \frac{π}{3}}) + R e s (φ, - e^{- i \frac{π}{3}})}$ $R e s (φ, e^{i \frac{π}{3}}) = \frac{1 + e^{i \frac{2 π}{3}}}{2 e^{i \frac{π}{3}} (2 i \frac{\sqrt{3}}{2}) (2 \frac{1}{2})}$ $= \frac{1}{2 i \sqrt{3}} (e^{- i \frac{π}{3}} + e^{i \frac{π}{3}}) = \frac{1}{2 i \sqrt{3}} 2 . \frac{1}{2} = \frac{1}{2 i \sqrt{3}}$ $R e s (φ, - e^{- i \frac{π}{3}}) = \frac{1 + e^{- i \frac{2 π}{3}}}{(- e^{- i \frac{π}{3}} - e^{i \frac{π}{3}}) (- e^{- i \frac{π}{3}} + e^{i \frac{π}{3}}) (- 2 e^{- i \frac{π}{3}})}$ $= \frac{1 + e^{- i \frac{2 π}{3}}}{2 (2 \frac{1}{2}) (2 i \frac{\sqrt{3}}{2}) e^{- i \frac{π}{3}}} = \frac{1}{2 i \sqrt{3}} (e^{i \frac{π}{3}} + e^{- i \frac{π}{3}})$ $= \frac{1}{2 i \sqrt{3}} (2 \frac{1}{2}) = \frac{1}{2 i \sqrt{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{2 i \sqrt{3}} + \frac{1}{2 i \sqrt{3}}} = 2 i π \frac{1}{i \sqrt{3}} = \frac{2 π}{\sqrt{3}}$ $2 I = \frac{2 π}{\sqrt{3}} \Rightarrow I = \frac{π}{\sqrt{3}} .$
	Answered by ajfour last updated on 13/May/18

	$= \int_{0}^{1} \frac{2 t^{2} d t}{t^{4} + t^{2} + 1}$ $= \int_{0}^{1} \frac{2 d t}{t^{2} + 1 + \frac{1}{t^{2}}}$ $= \int_{0}^{1} \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - 1} + \int_{0}^{1} \frac{(1 + \frac{1}{t^{2}}) d t}{3 + {(t - \frac{1}{t})}^{2}}$ $= \int_{\infty}^{2} \frac{d z}{z^{2} - 1} + \int_{- \infty}^{0} \frac{d z}{z^{2} + 3}$ $= \frac{1}{2} \ln (\frac{z - 1}{z + 1}) ∣_{\infty}^{2} + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{z}{\sqrt{3}}) ∣_{- \infty}^{0}$ $= \frac{1}{2} \ln (\frac{1}{3}) + \frac{π}{2 \sqrt{3}}$ $o r = \frac{π}{2 \sqrt{3}} - \frac{1}{2} \ln 3 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18

	$x = t^{2}$ $d x = 2 t d t$ $\int_{0}^{1} \frac{t \times 2 t d t}{t^{4} + t^{2} + 1}$ $= \int_{0}^{1} \frac{2}{t^{2} + 1 + \frac{1}{t^{2}}} d t$ $= \int_{0}^{1} \frac{(1 - 1 / t^{2}) + (1 + 1 / t^{2})}{t^{2} + \frac{1}{t^{2}} + 1} d t$ $= \int_{0}^{1} \frac{1 - 1 / t^{2}}{{(t + 1 / t)}^{2} - 2 + 1} d t + \int_{0}^{1} \frac{1 + 1 / t^{2}}{{(t - 1 / t)}^{2}} d$ $z_{1 = t + 1 / t s o l i m i t}$ $\int_{\infty}^{2} \frac{{d z}_{1}}{z_{1}^{2} - 1} + \int_{- \infty}^{0} \frac{{d z}_{2}}{z_{2}^{2} + 3}$ $\frac{1}{2} \int_{\infty}^{2} \frac{(z_{1} + 1) - (z_{1} - 1)}{(z_{1} + 1) (z_{1} - 1)} {d z}_{1} + \frac{1}{\sqrt{3}} ∣ {t a n}^{- 1} (\frac{z_{2}}{\sqrt{3}}) ∣_{- \infty}^{0}$ $= \frac{1}{2} ∣ l n (\frac{z_{1} - 1}{z_{1} + 1}) ∣_{\infty}^{2} + ∣ \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{z_{2}}{\sqrt{3}}) ∣_{- \infty}^{0}$ $\frac{1}{2} l n (\frac{1}{3}) - \frac{1}{2} ∣ l n (\frac{1 - \frac{1}{\infty}}{1 + \frac{1}{\infty}}) ∣ + \frac{1}{\sqrt{3}} (0 + Π / 2)$ $= - \frac{1}{2} l n 3 + Π / 2 \sqrt{3}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/May/18

	$t r u e . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/May/18

	$i n h u r r y f o r g o t t o c h a n g e . . . t h a n x . . .$
	Answered by MJS last updated on 13/May/18

	$\int \frac{\sqrt{x}}{x^{2} + x + 1} d x =$ $[u = \sqrt{x} \to d x = 2 \sqrt{x} d u]$ $= 2 \int \frac{u^{2}}{u^{4} + u^{2} + 1} d u = 2 \int \frac{u^{2}}{(u^{2} - u + 1) (u^{2} + u + 1)} d u =$ $= 2 \int (\frac{u}{2 (u^{2} - u + 1)} - \frac{u}{2 (u^{2} + u + 1)}) d u =$ $= \int \frac{u}{u^{2} - u + 1} d u - \int \frac{u}{u^{2} + u + 1} d u =$ $= \int (\frac{2 u - 1}{2 (u^{2} - u + 1)} + \frac{1}{2 (u^{2} - u + 1)}) d u - \int (\frac{2 u + 1}{2 (u^{2} + u + 1)} - \frac{1}{2 (u^{2} + u + 1)}) d u =$ $= \frac{1}{2} (\int \frac{2 u - 1}{u^{2} - u + 1} d u + \int \frac{1}{u^{2} - u + 1} d u - \int \frac{2 u + 1}{u^{2} + u + 1} d u + \int \frac{1}{u^{2} + u + 1} d u) =$ $\int \frac{2 u \pm 1}{u^{2} \pm u + 1} d u =$ $[v = u^{2} \pm u + 1 \to d u = \frac{d v}{2 u \pm 1}]$ $= \int \frac{1}{v} d v = \ln ∣ v ∣= \ln ∣ u^{2} \pm u + 1 ∣$ $\int \frac{1}{u^{2} \pm u + 1} d u = \int \frac{1}{{(u \pm \frac{1}{2})}^{2} + \frac{3}{4}} d u =$ $[v = \frac{\sqrt{3}}{3} (2 u \pm 1) \to d u = \frac{\sqrt{3}}{2} d v]$ $\frac{2 \sqrt{3}}{3} \int \frac{1}{v^{2} + 1} d v = \frac{2 \sqrt{3}}{3} \arctan v = \frac{2 \sqrt{3}}{3} \arctan \frac{\sqrt{3}}{3} (2 u \pm 1)$ $= \frac{1}{2} (\ln ∣ u^{2} - u + 1 ∣ + \frac{2 \sqrt{3}}{3} \arctan \frac{\sqrt{3}}{3} (2 u - 1) - \ln ∣ u^{2} + u + 1 ∣ + \frac{2 \sqrt{3}}{3} \arctan \frac{\sqrt{3}}{3} (2 u + 1)) =$ $= \frac{\ln ∣ x - \sqrt{x} + 1 ∣ - \ln ∣ x + \sqrt{x} + 1 ∣}{2} + \frac{\sqrt{3}}{3} (\arctan \frac{\sqrt{3}}{3} (2 \sqrt{x} - 1) + arcran \frac{\sqrt{3}}{3} (2 \sqrt{x} + 1))$ $\underset{0}{\int^{1}} \frac{\sqrt{x}}{x^{2} + x + 1} d x = \frac{\sqrt{3}}{6} π - \frac{1}{2} \ln 3$
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