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Question Number 148000 by vvvv last updated on 25/Jul/21
∫+∞−∞dx(x2+k2)32
Answered by gsk2684 last updated on 25/Jul/21
2∫∞x=0dxx3(1+k2x2)32changement1+k2x2=t⇒−2k2x3dx=dt2∫1t=∞t−32dt−2k2=−1k2[t−32+1−32+1]∞1=−1k2[t−12−12]∞1=2k2[1t]∞1=2k2[1−0]=2k2
Answered by mathmax by abdo last updated on 25/Jul/21
Uk=∫−∞+∞dx(x2+k2)32changementx=ktanθgiveUk=∫−π2π2k(1+tan2θ)k3(1+tan2θ)32dθ=1k2∫−π2π2dθ1+tan2θ=1k2∫−π2π2cosθdθ=1k2[sinθ]−π2π2=1k2(1−(−1))=2k2(k≠0)
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