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Question Number 104197 by mathmax by abdo last updated on 20/Jul/20

calculate ∫_5 ^(+∞)  (dx/((x^2 −9)^4 ))

calculate5+dx(x29)4

Answered by Dwaipayan Shikari last updated on 20/Jul/20

∫_5 ^(+∞) ((2xdx)/(2x(x^2 −9)^4 ))=(1/2)∫_(16) ^(+∞) (dt/(√(9+t))).(1/t^4 )                          {t=9tan^2 θ, 1=18tanθsec^2 θ(dθ/dt)  (1/6)∫_(tan^(−1) (4/3)) ^(π/2) ((18tanθsec^2 θdθ)/(secθ.9^4 tan^8 θ))   {take x^2 −9=t  (1/3).(1/9^3 )∫_(tan^(−1) (4/3)) ^(π/2) ((secθ)/(tan^7 θ))dθ.....continue

5+2xdx2x(x29)4=1216+dt9+t.1t4{t=9tan2θ,1=18tanθsec2θdθdt16tan143π218tanθsec2θdθsecθ.94tan8θ{takex29=t13.193tan143π2secθtan7θdθ.....continue

Answered by mathmax by abdo last updated on 20/Jul/20

A =∫_5 ^∞  (dx/((x^2 −9)^4 )) ⇒A =∫_5 ^∞  (dx/((x−3)^4 (x+3)^4 )) =∫_5 ^∞  (dx/((((x−3)/(x+3)))^4 (x+3)^8 ))  we do the changement ((x−3)/(x+3)) =t ⇒x−3 =tx+3t ⇒(1−t)x =3t+3 ⇒x =((3t+3)/(1−t))  (dx/dx) =((3(1−t)+(3t+3))/((1−t)^2 )) =(6/((1−t)^2 ))  x+3 =((3t+3)/(1−t)) +3 =((3t+3+3−3t)/(1−t)) =(6/(1−t)) ⇒A =∫_(1/4) ^1  (6/((1−t)^2 t^4 ((6/(1−t)))^8 ))dt  =(1/6^7 ) ∫_(1/4) ^1   (((t−1)^8 )/((t−1)^2  t^4 ))dt =(1/6^7 ) ∫_(1/4) ^1   (((t−1)^6 )/t^4 ) dt   =(1/6^7 ) ∫_(1/4) ^1  ((Σ_(k=0) ^6  C_6 ^(k )  t^k )/t^4 )dt =(1/6^7 ) Σ_(k=0) ^6  C_6 ^k  ∫_(1/4) ^1  t^(k−4)  dt  =(1/6^7 ) Σ_(k=0and k≠3) ^6  C_6 ^k  [(1/(k−3))t^(k−3) ]_(1/4) ^1    +(1/6^7 )C_6 ^3  [lnt]_(1/4) ^1   A =(1/6^7 ) Σ_(k=0 and k≠3) ^6  (C_6 ^k /(k−3)){1−(1/4^(k−3) )} +(C_6 ^3 /6^7 )×(2ln2)

A=5dx(x29)4A=5dx(x3)4(x+3)4=5dx(x3x+3)4(x+3)8wedothechangementx3x+3=tx3=tx+3t(1t)x=3t+3x=3t+31tdxdx=3(1t)+(3t+3)(1t)2=6(1t)2x+3=3t+31t+3=3t+3+33t1t=61tA=1416(1t)2t4(61t)8dt=167141(t1)8(t1)2t4dt=167141(t1)6t4dt=167141k=06C6ktkt4dt=167k=06C6k141tk4dt=167k=0andk36C6k[1k3tk3]141+167C63[lnt]141A=167k=0andk36C6kk3{114k3}+C6367×(2ln2)

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