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Question Number 148000 by vvvv last updated on 25/Jul/21

∫_(−∞) ^(+∞) (dx/((x^2 +k^2 )^(3/2) ))

+dx(x2+k2)32

Answered by gsk2684 last updated on 25/Jul/21

2∫_(x=0) ^∞ (dx/(x^3 (1+(k^2 /x^2 ))^(3/2) ))   changement 1+(k^2 /x^2 )=t ⇒ −((2k^2 )/x^3 )dx=dt  2∫_(t=∞) ^1 t^(−(3/2))  (dt/(−2k^2 ))=((−1)/k^2 )[(t^(((−3)/2)+1) /(−(3/2)+1))]_∞ ^1   =((−1)/k^2 )[(t^(−(1/2)) /(−(1/2)))]_∞ ^1 =(2/k^2 )[(1/( (√t)))]_∞ ^1 =(2/k^2 )[1−0]=(2/k^2 )

2x=0dxx3(1+k2x2)32changement1+k2x2=t2k2x3dx=dt21t=t32dt2k2=1k2[t32+132+1]1=1k2[t1212]1=2k2[1t]1=2k2[10]=2k2

Answered by mathmax by abdo last updated on 25/Jul/21

U_k =∫_(−∞) ^(+∞)  (dx/((x^2  +k^2 )^(3/2) ))  changement x=ktanθ give  U_k =∫_(−(π/2)) ^(π/2)  ((k(1+tan^2 θ))/(k^3 (1+tan^2 θ)^(3/2) ))dθ =(1/k^2 )∫_(−(π/2)) ^(π/2)  (dθ/( (√(1+tan^2 θ))))  =(1/k^2 )∫_(−(π/2)) ^(π/2) cosθ dθ =(1/k^2 )[sinθ]_(−(π/2)) ^(π/2)  =(1/k^2 )(1−(−1)) =(2/k^2 )    (k≠0)

Uk=+dx(x2+k2)32changementx=ktanθgiveUk=π2π2k(1+tan2θ)k3(1+tan2θ)32dθ=1k2π2π2dθ1+tan2θ=1k2π2π2cosθdθ=1k2[sinθ]π2π2=1k2(1(1))=2k2(k0)

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