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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?

changementx=tgiveI=01tt4+t2+12tdt=201t2t4+t2+1dtch.t=1uI=21+1u2(1u4+1u2+1)(duu2)=21+duu4+u2+12I=2{01t2t4+t2+1dt+1+dtt4+t2+1}I=01+t2t4+t2+1dt2I=+1+t2t4+t2+1dtletconsiderthecomplexfunctionφ(z)=1+z2z4+z2+1polesofφ?

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)) .

letsolvez4+z2+1=0changementz2=tt2+t+1=0therootst1=ei2π3,t2=ei2π3φ(z)=1+z2(z2ei2π3)(z2ei2π3)=1+z2(zeiπ3)(z+eiπ3)(zeiπ3)(z+eiπ3)+φ(z)dz=2iπ{Res(φ,eiπ3)+Res(φ,eiπ3)}Res(φ,eiπ3)=1+ei2π32eiπ3(2i32)(212)=12i3(eiπ3+eiπ3)=12i32.12=12i3Res(φ,eiπ3)=1+ei2π3(eiπ3eiπ3)(eiπ3+eiπ3)(2eiπ3)=1+ei2π32(212)(2i32)eiπ3=12i3(eiπ3+eiπ3)=12i3(212)=12i3+φ(z)dz=2iπ{12i3+12i3}=2iπ1i3=2π32I=2π3I=π3.

Answered by ajfour last updated on 13/May/18

  =∫_0 ^(  1) ((2t^2 dt)/(t^4 +t^2 +1))    =∫_0 ^(  1) ((2dt)/(t^2 +1+(1/t^2 )))      =∫_0 ^(  1) (((1−(1/t^2 ))dt)/((t+(1/t))^2 −1)) + ∫_0 ^(  1) (((1+(1/t^2 ))dt)/(3+(t−(1/t))^2 ))  =∫_∞ ^(  2) (dz/(z^2 −1)) +∫_(−∞) ^(  0) (dz/(z^2 +3))    =(1/2)ln (((z−1)/(z+1)))∣_∞ ^2 +(1/(√3))tan^(−1) ((z/(√3)))∣_(−∞) ^0   =(1/2)ln ((1/3))+(π/(2(√3)))    or    = (π/(2(√3)))−(1/2)ln 3   .

=012t2dtt4+t2+1=012dtt2+1+1t2=01(11t2)dt(t+1t)21+01(1+1t2)dt3+(t1t)2=2dzz21+0dzz2+3=12ln(z1z+1)2+13tan1(z3)0=12ln(13)+π23or=π2312ln3.

Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18

x=t^2   dx=2tdt  ∫_0 ^1 ((t×2tdt)/(t^4 +t^2 +1))  =∫_0 ^1 (2/(t^2 +1+(1/t^2 )))dt  =∫_0 ^1 (((1−1/t^2 ) +(1+1/t^2 ))/(t^2 +(1/t^(2 ) )+1))dt  =∫_0 ^1 ((1−1/t^2 )/((t+1/t)^2 −2+1))dt+∫_0 ^1 ((1+1/t^2 )/((t−1/t)^2 ))d  z_(1=t+1/t so limit)      ∫_∞ ^2 (dz_1 /(z_1 ^2 −1))+∫_(−∞) ^0 (dz_2 /(z_2 ^2 +3))  (1/2)∫_∞ ^2 (((z_1 +1)−(z_1 −1))/((z_1 +1)(z_1 −1)))dz_1 +(1/(√3))∣tan^(−1) ((z_2 /(√3)))∣_(−∞) ^0   =(1/2)∣ln(((z_1 −1)/(z_1 +1)))∣_∞ ^2 +∣(1/(√3))tan^(−1) ((z_2 /(√3)))∣_(−∞) ^0   (1/2)ln((1/3))−(1/2)∣ln(((1−(1/∞))/(1+(1/∞))))∣+(1/(√3))(0+Π/2)  =−(1/2)ln3+Π/2(√3)

x=t2dx=2tdt01t×2tdtt4+t2+1=012t2+1+1t2dt=01(11/t2)+(1+1/t2)t2+1t2+1dt=0111/t2(t+1/t)22+1dt+011+1/t2(t1/t)2dz1=t+1/tsolimit2dz1z121+0dz2z22+3122(z1+1)(z11)(z1+1)(z11)dz1+13tan1(z23)0=12ln(z11z1+1)2+13tan1(z23)012ln(13)12ln(111+1)+13(0+Π/2)=12ln3+Π/23

Commented by tanmay.chaudhury50@gmail.com last updated on 13/May/18

true...

true...

Commented by tanmay.chaudhury50@gmail.com last updated on 13/May/18

in hurry forgot to change...thanx...

inhurryforgottochange...thanx...

Answered by MJS last updated on 13/May/18

∫((√x)/(x^2 +x+1))dx=            [u=(√x) → dx=2(√x)du]  =2∫(u^2 /(u^4 +u^2 +1))du=2∫(u^2 /((u^2 −u+1)(u^2 +u+1)))du=  =2∫((u/(2(u^2 −u+1)))−(u/(2(u^2 +u+1))))du=  =∫(u/(u^2 −u+1))du−∫(u/(u^2 +u+1))du=  =∫(((2u−1)/(2(u^2 −u+1)))+(1/(2(u^2 −u+1))))du−∫(((2u+1)/(2(u^2 +u+1)))−(1/(2(u^2 +u+1))))du=  =(1/2)(∫((2u−1)/(u^2 −u+1))du+∫(1/(u^2 −u+1))du−∫((2u+1)/(u^2 +u+1))du+∫(1/(u^2 +u+1))du)=            ∫((2u±1)/(u^2 ±u+1))du=                      [v=u^2 ±u+1 → du=(dv/(2u±1))]            =∫(1/v)dv=ln ∣v∣=ln ∣u^2 ±u+1∣            ∫(1/(u^2 ±u+1))du=∫(1/((u±(1/2))^2 +(3/4)))du=                      [v=((√3)/3)(2u±1) → du=((√3)/2)dv]            ((2(√3))/3)∫(1/(v^2 +1))dv=((2(√3))/3)arctan v=((2(√3))/3)arctan ((√3)/3)(2u±1)  =(1/2)(ln ∣u^2 −u+1∣+((2(√3))/3)arctan ((√3)/3)(2u−1)−ln ∣u^2 +u+1∣+((2(√3))/3)arctan ((√3)/3)(2u+1))=  =((ln ∣x−(√x)+1∣−ln ∣x+(√x)+1∣)/2)+((√3)/3)(arctan ((√3)/3)(2(√x)−1)+arcran ((√3)/3)(2(√x)+1))  ∫_0 ^1 ((√x)/(x^2 +x+1))dx=((√3)/6)π−(1/2)ln 3

xx2+x+1dx=[u=xdx=2xdu]=2u2u4+u2+1du=2u2(u2u+1)(u2+u+1)du==2(u2(u2u+1)u2(u2+u+1))du==uu2u+1duuu2+u+1du==(2u12(u2u+1)+12(u2u+1))du(2u+12(u2+u+1)12(u2+u+1))du==12(2u1u2u+1du+1u2u+1du2u+1u2+u+1du+1u2+u+1du)=2u±1u2±u+1du=[v=u2±u+1du=dv2u±1]=1vdv=lnv∣=lnu2±u+11u2±u+1du=1(u±12)2+34du=[v=33(2u±1)du=32dv]2331v2+1dv=233arctanv=233arctan33(2u±1)=12(lnu2u+1+233arctan33(2u1)lnu2+u+1+233arctan33(2u+1))==lnxx+1lnx+x+12+33(arctan33(2x1)+arcran33(2x+1))10xx2+x+1dx=36π12ln3

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