Question Number 103742 by mathmax by abdo last updated on 17/Jul/20

calculate ∫_0 ^∞     (dx/((2x+1)^4 (x+3)^5 ))

Answered by mathmax by abdo last updated on 17/Jul/20

A=∫_0 ^∞     (dx/((2x+1)^4 (x+3)^5 )) ⇒A =∫_0 ^∞   (dx/((((2x+1)/(x+3)))^4 (x+3)^9 ))   we do the changement  ((2x+1)/(x+3)) =t ⇒2x+1 =tx +3t ⇒(2−t)x =3t−1 ⇒x =((3t−1)/(2−t)) ⇒  (dx/dt)  =((3(2−t)+3t−1)/((2−t)^2 )) =(5/((2−t)^2 )) and x+3 =((3t−1)/(2−t)) +3 =((3t−1+6−3t)/(2−t)) =(5/(2−t))  ⇒A =∫_(1/3) ^2    (5/((2−t)^2  t^4 ((5/(2−t)))^9 )) dt =(1/5^8 ) ∫_(1/3) ^2  (((2−t)^9 )/((2−t)^2  t^4 ))dt  =−(1/5^8 ) ∫_(1/3) ^2  (((t−2)^7 )/t^4 )dt =−(1/5^8 ) ∫_(1/3) ^2  ((Σ_(k=0) ^(7 )  C_7 ^k  t^k (−2)^(7−k) )/t^4 )dt  =−(1/5^8 ) Σ_(k=0) ^7  C_7 ^k (−2)^(7−k)  ∫_(1/3) ^2  t^(k−4 ) dt  =−(1/5^8 ){ Σ_(k=0and k≠3) ^7  (−2)^(7−k)  C_7 ^k  [(1/(k−3))t^(k−3) ]_(1/3) ^2   +(−2)^4  C_7 ^3  [ln∣t∣]_(1/3) ^2 }  A=−(1/5^8 ){ Σ_(k=0 and k≠3) ^7  (−2)^(7−k )  (C_7 ^k /(k−3)){2^(k−3) −((1/3))^(k−3) }+(−2)^4  C_7 ^3 (ln2+ln3))}