Question Number 103591 by mathmax by abdo last updated on 16/Jul/20

calculate  ∫_3 ^(+∞)       (dx/((x^2 −1)^3 (x+2)^2 ))

Commented byWorm_Tail last updated on 16/Jul/20

       ∫_3 ^(+∞)       (dx/((x^2 −1)^3 (x+2)^2 ))     pfd        ∫_3 ^(oo) ((−3)/(16(x+1)))+(4/(27(x+2)))+(1/(16(x+1)^2 ))+(1/(27(x+2)^2 ))−(1/(8(x+1)^3 ))+((17)/(432(x−1)))−((13)/(432(x−1)^2 ))+(1/(72(x−1)^3 ))dx        [((−3ln(x+1))/(16))+(4/(27))ln(x+2)−(1/(16(x+1)))−(1/(27(x+2)))+(1/(16(x+1)^2 ))+((17ln(x−1))/(432))+((13)/(432(x−1)))−(1/(144(x−1)^2 ))]_3 ^(oo)        [((ln(((x+2)^(64) (x−1)^(17) )/((x+1)^(81) )))/(432))−(1/(27(x+2)))+(1/(16(x+1)^2 ))+((13)/(432(x−1)))−(1/(144(x−1)^2 ))]_3 ^(oo)    lim_(t→oo)     [((ln(((t+2)^(64) (t−1)^(17) )/((t+1)^(81) )))/(432))−(1/(27(t+2)))+(1/(16(t+1)^2 ))+((13)/(432(t−1)))−(1/(144(t−1)^2 ))]   −    [((ln(((3+2)^(64) (3−1)^(17) )/((3+1)^(81) )))/(432))−(1/(27(3+2)))+(1/(16(3+1)^2 ))+((13)/(432(3−1)))−(1/(144(3−1)^2 ))]      lim_(t→oo)     [((ln(1))/(432))−0+0+0−0]−[((ln((5^(64) ×2^(17) )/4^(81) ))/(432))+((113)/(11520))]   −[((64ln(5)−145ln(2))/(432))+((113)/(11520))]   ((−64ln(5)+145ln(2))/(432))−((113)/(11520))

Commented byWorm_Tail last updated on 16/Jul/20

mistake  might   have   crept   in

Commented byabdomathmax last updated on 16/Jul/20

can you chow how do you find the decomposition?

Commented byWorm_Tail last updated on 16/Jul/20

       ∫_3 ^(+∞)       (dx/((x^2 −1)^3 (x+2)^2 ))     pfd        (A/((x+1)))+(B/((x+2)))+(C/((x+1)^2 ))+(D/((x+2)^2 ))+(E/((x+1)^3 ))+(F/((x−1)))+(G/((x−1)^2 ))+(H/((x−1)^3 ))=(1/((x^2 −1)^3 (x+2)^2 ))  you continue  from here

Answered by abdomathmax last updated on 16/Jul/20

I =∫_3 ^∞  (dx/((x^2 −1)^3 (x+2)^2 )) ⇒ I =∫_3 ^∞  (dx/((x−1)^3 (x+1)^3 (x+2)^2 ))  =∫_3 ^∞  (dx/((((x−1)/(x+1)))^3 (x+1)^6 (x+2)^2 )) we do tbe cha7gement  ((x−1)/(x+1)) =t ⇒x−1 =tx+t ⇒(1−t)x =t+1 ⇒  x =((1+t)/(1−t)) ⇒ (dx/dt) =((1−t+(1+t))/((1−t)^2 )) =(2/((1−t)^2 ))  x+1 =((1+t)/(1−t)) +1 =((1+t+1−t)/(1−t)) =(2/(1−t))  x+2 =((1+t)/(1−t))+2 =((1+t+2−2t)/(1−t)) =((3−t)/(1−t)) ⇒  I = ∫_(1/2) ^1   ((2dt)/((1−t)^2 t^3 ((2/(1−t)))^6 (((3−t)/(1−t)))^2 ))  =2∫_(1/2) ^1    (((t−1)^8 )/((t−1)^2 t^3  .2^6 (3−t)^2 ))dt  =(1/2^5 ) ∫_(1/2) ^1    (((t−1)^6 )/(t^3 (3−t)^2 ))dt =(1/2^5 ) ∫_(1/2) ^1  (((t−1)^6 )/(((t/(t−3)))^3 (t−3)^5 ))dt  again ch.(t/(t−3)) =z ⇒t  =zt−3z ⇒(1−z)t =−3z ⇒  t =((−3z)/(1−z)) =((3z)/(z−1)) and t−3 =((3z)/(z−1))−3 =((3z−3z+3)/(z−1))  =(3/(z−1))  and t−1 =((3z)/(z−1))−1 =((3z−z+1)/(z−1)) =((2z+1)/(z−1))  I =(1/2^5 )∫_(−(1/5)) ^(−(1/2))   (((((2z+1)/(z−1)))^6 )/(z^3 ((3/(z−1)))^5 ))×((−3)/((z−1)^2 ))dz  =−(1/(2^5 .3^4 )) ∫_(−(1/5)) ^(−(1/2))    (((2z+1)^6 (z−1)^5 )/(z^3 (z−1)^8 )) dz  =−(1/(2^5 .3^4 )) ∫_(−(1/5)) ^(−(1/2))   (((2z+1)^6 )/(z^3 (z−1)^3 ))dz  =−(1/(2^5 .3^4 )) ∫_(−(1/5)) ^(−(1/2))  ((Σ_(k=0) ^6  C_6 ^k (2z)^k )/(z^3 (z−1)^3 ))dz  =−(1/(2^5 .3^4 )) ∫_(−(1/5)) ^(−(1/2)) Σ_(k=0) ^6  2^k  C_6 ^k  (z^k /(z^3 (z−1)^3 ))dz  =−(1/(2^5 .3^4 )) Σ_(k=0) ^6  2^k  C_6 ^k  ∫_(−(1/5)) ^(−(1/2))  (z^k /(z^3 (z−1)^3 ))dz  after we decompose F_k (z) =(z^k /(z^3 (z−1)^3 ))  ...be continued...