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Question Number 133537 by mathmax by abdo last updated on 22/Feb/21

calculate ∫_1 ^∞  (dx/((x^2 +x+1)^9 ))

calculate1dx(x2+x+1)9

Answered by Ñï= last updated on 23/Feb/21

∫_1 ^∞ (dx/((1+x+x^2 )^9 ))  =∫_1 ^∞ (dx/([(x+(1/2))^2 +(3/4)]^9 ))  =(√(3/4))∫_(π/3) ^(π/2) ((sec^2 θdθ)/(((3/4)tan^2 θ+(3/4))^9 ))  =((√(3/4))/( ((3/4))^9 ))∫_(π/3) ^(π/2) cos^(16) θdθ  =((3/4))^(−17/2) Re(∫_(π/3) ^(π/2) e^(16iθ) dθ)  =((3/4))^(−17/2) Re((1/(16i))(e^(8iπ) −e^(16iπ/3) )  =(2^(12) /3^8 )

1dx(1+x+x2)9=1dx[(x+12)2+34]9=34π3π2sec2θdθ(34tan2θ+34)9=34(34)9π/3π/2cos16θdθ=(34)17/2Re(π/3π/2e16iθdθ)=(34)17/2Re(116i(e8iπe16iπ/3)=21238

Answered by mathmax by abdo last updated on 24/Feb/21

Φ=∫_1 ^∞  (dx/((x^2  +x+1)^9 )) ⇒Φ =∫_1 ^(∞ )  (dx/(((x+(1/2))^2  +(3/4))^9 ))  =_(x+(1/2)=((√3)/2)t)     ∫_(√3) ^∞  ((4/3))^9  (1/((t^2 +1)^9 ))((√3)/2)dt =((√3)/2)((4/3))^9  ∫_(√3) ^∞ (dt/((t^2  +1)^9 ))  wehave  ∫_(√3) ^∞  (dt/((t^2  +1)^9 )) =∫_(√3) ^∞  (dt/((t−i)^9 (t+i)^9 ))=∫_(√3) ^∞  (dt/((((t−i)/(t+i)))^9 (t+i)^(18) ))  changement ((t−i)/(t+i))=z give t−i=zt+iz ⇒(1−z)t=i(1+z) ⇒  t=i((1+iz)/(1−iz)) ⇒(dt/dz)=i((i(1−iz)−(1+iz)(−i))/((1−iz)^2 ))=i((i+z+i−z)/((1−iz)^2 ))=((−2)/((1−iz)^2 ))  and t+i =((i−z)/(1−iz))+i =((i−z+i+z)/((1−iz)))=((2i)/((1−iz))) ⇒  Φ =∫_(((√3)−i)/( (√3)+i)) ^1   ((−2)/((1−iz)^2 z^9 (((2i)/(1−iz)))^(18) ))dz  =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i))      (((1−iz)^(16) )/z^9 )dz =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i))  ((Σ_(k=0) ^(16)  C_(16) ^k (iz)^k (−1)^(16−k) )/z^9 )dz  =−(2/(2^(18)  )) ∫_1 ^(((√3)−i)/( (√3) +i))  Σ_(k=0) ^(16)  (−1)^k  C_(16) ^k  i^k  z^(k−9)  dz  =−(1/2^(17) )Σ_(k=0 ,k≠8) ^(16)  (−i)^k  C_(16) ^k  [(1/(k−8))z^(k−8) ]_1 ^(((√3)−i)/( (√(3 ))+i))   −(1/2^(17) )C_(16) ^8  [lnz]_1 ^(((√3)−i)/( (√3)+i))   =−(1/2^(17) )Σ_(k=0 and k≠8) ^(16)    (((−1)^k  C_(16) ^k )/(k−8)){((((√3)−i)/( (√3)+i)))^(k−8) −1}  −(1/2^(17) )C_(16) ^8  ln((((√3)−i)/( (√3)+i)))

Φ=1dx(x2+x+1)9Φ=1dx((x+12)2+34)9=x+12=32t3(43)91(t2+1)932dt=32(43)93dt(t2+1)9wehave3dt(t2+1)9=3dt(ti)9(t+i)9=3dt(tit+i)9(t+i)18changementtit+i=zgiveti=zt+iz(1z)t=i(1+z)t=i1+iz1izdtdz=ii(1iz)(1+iz)(i)(1iz)2=ii+z+iz(1iz)2=2(1iz)2andt+i=iz1iz+i=iz+i+z(1iz)=2i(1iz)Φ=3i3+i12(1iz)2z9(2i1iz)18dz=2(2i)1813i3+i(1iz)16z9dz=2(2i)1813i3+ik=016C16k(iz)k(1)16kz9dz=221813i3+ik=016(1)kC16kikzk9dz=1217k=0,k816(i)kC16k[1k8zk8]13i3+i1217C168[lnz]13i3+i=1217k=0andk816(1)kC16kk8{(3i3+i)k81}1217C168ln(3i3+i)

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