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

calculate ∫_5 ^(+∞)  (dx/((x^2 −9)^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

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)