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Question Number 133537 by mathmax by abdo last updated on 22/Feb/21
calculate∫1∞dx(x2+x+1)9
Answered by Ñï= last updated on 23/Feb/21
∫1∞dx(1+x+x2)9=∫1∞dx[(x+12)2+34]9=34∫π3π2sec2θdθ(34tan2θ+34)9=34(34)9∫π/3π/2cos16θdθ=(34)−17/2Re(∫π/3π/2e16iθdθ)=(34)−17/2Re(116i(e8iπ−e16iπ/3)=21238
Answered by mathmax by abdo last updated on 24/Feb/21
Φ=∫1∞dx(x2+x+1)9⇒Φ=∫1∞dx((x+12)2+34)9=x+12=32t∫3∞(43)91(t2+1)932dt=32(43)9∫3∞dt(t2+1)9wehave∫3∞dt(t2+1)9=∫3∞dt(t−i)9(t+i)9=∫3∞dt(t−it+i)9(t+i)18changementt−it+i=zgivet−i=zt+iz⇒(1−z)t=i(1+z)⇒t=i1+iz1−iz⇒dtdz=ii(1−iz)−(1+iz)(−i)(1−iz)2=ii+z+i−z(1−iz)2=−2(1−iz)2andt+i=i−z1−iz+i=i−z+i+z(1−iz)=2i(1−iz)⇒Φ=∫3−i3+i1−2(1−iz)2z9(2i1−iz)18dz=2(2i)18∫13−i3+i(1−iz)16z9dz=2(2i)18∫13−i3+i∑k=016C16k(iz)k(−1)16−kz9dz=−2218∫13−i3+i∑k=016(−1)kC16kikzk−9dz=−1217∑k=0,k≠816(−i)kC16k[1k−8zk−8]13−i3+i−1217C168[lnz]13−i3+i=−1217∑k=0andk≠816(−1)kC16kk−8{(3−i3+i)k−8−1}−1217C168ln(3−i3+i)
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