Question Number 101269 by mathmax by abdo last updated on 01/Jul/20

calculate ∫_4 ^(+∞) (dx/((x−2)^5 (x+3)^7 ))

Answered by mathmax by abdo last updated on 05/Jul/20

I =∫_4 ^(∞ ) (dx/((x−2)^5 (x+3)^7 )) ⇒ I =∫_4 ^∞ (dx/((((x+3)/(x−2)))^7 (x−2)^(12) )) we do the cha7gement ((x+3)/(x−2)) =t ⇒x+3 =tx−2t ⇒(1−t)x =−2t−3 ⇒x =((2t+3)/(t−1)) ⇒ (dx/dt) =((2(t−1)−(2t+3))/((t−1)^2 )) =((−5)/((t−1)^2 )) and x−2 =((2t+3)/(t−1))−2 =((2t+3−2t+2)/(t−1)) =(5/(t−1)) ⇒ I =∫_(7/2) ^1 ((−5dt)/((t−1)^2 .t^7 .((5/(t−1)))^(12) )) =(1/5^(11) ) ∫_1 ^(7/2) (((t−1)^(12) )/((t−1)^2 t^7 ))dt ⇒ 5^(11) ×I =∫_1 ^(7/2) (((t−1)^(10) )/t^7 )dt =∫_1 ^(7/2) ((Σ_(k=0) ^(10 ) C_(10) ^k t^k (−1)^(10−k) )/t^7 )dt =Σ_(k=0) ^(10) (−1)^(k ) C_(10) ^k ∫_1 ^(7/(2 )) t^(k−7) dt =Σ_(k=0 and k≠6) ^(10) (−1)^k C_(10) ^k [(1/(k−6))t^(k−6) ]_1 ^(7/2) +C_(10) ^6 [ln∣t∣]_1 ^(7/2) =Σ_(k=0 and k≠6) ^(10) (−1)^k (C_(10) ^k /(k−6)){((7/2))^(k−6) −1} +C_(10) ^6 ln((7/2)) ⇒ I =(1/5^(11) ) Σ_(k=0 and k≠6) ^(10) (((−1)^k C_(10) ^k )/(k−6)){ ((7/2))^(k−6) −1}+(C_(10) ^6 /5^(11) )ln((7/2)) .