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Question Number 124531 by Bird last updated on 03/Dec/20
calculate ∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 ))
calculate∫0∞dx(2x+1)4(x+3)5
Answered by mathmax by abdo last updated on 04/Dec/20
let I =∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 )) ⇒I =∫_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)(−1))/((2−t)^2 ))=((6−3t+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)) ⇒ I =∫_(1/3) ^2 (5/((2−t)^2 t^4 ((5/(2−t)))^9 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^9 )/(t^4 ((2−t)^2 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^7 )/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_(k0) ^7 (−2)^(7−k) C_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 [ln2−ln((1/3))) I=−(1/5^8 )Σ_(k=0 and ≠3) ^7 (−2)^(7−k) (C_7 ^k /(k−3)){2^(k−3) −(1/3^(k−3) )} +16 C_7 ^3 ln(6)
letI=∫0∞dx(2x+1)4(x+3)5⇒I=∫0∞dx(2x+1x+3)4(x+3)9wedothechangement2x+1x+3=t⇒2x+1=tx+3t⇒(2−t)x=3t−1⇒x=3t−12−t⇒dxdt=3(2−t)−(3t−1)(−1)(2−t)2=6−3t+3t−1(2−t)2=5(2−t)2andx+3=3t−12−t+3=3t−1+6−3t2−t=52−t⇒I=∫1325(2−t)2t4(52−t)9dt=158∫132(2−t)9t4((2−t)2dt=158∫132(2−t)7t4dt=−158∫132(t−2)7t4dt=−158∫132∑k=07C7ktk(−2)7−kt4dt=−158∑k07(−2)7−kC7k∫132tk−4dt=−158{∑k=0andk≠37(−2)7−kC7k[1k−3tk−3]132+(−2)4C73[ln2−ln(13))I=−158∑k=0and≠37(−2)7−kC7kk−3{2k−3−13k−3}+16C73ln(6)

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