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


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Question Number 146196 by mathmax by abdo last updated on 11/Jul/21

$calculate \int_{0}^{\infty} \frac{dx}{{(2 x + 1)}^{4} {(x + 3)}^{5}}$
	Answered by mathmax by abdo last updated on 12/Jul/21
	$Ψ=∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 )) ⇒Ψ=∫_0 ^∞ (dx/((((2x+1)/(x+3)))^4 (x+3)^9 )) we do the changement ((2x+1)/(x+3))=t ⇒2x+1=tx+3t ⇒(2−t)x=3t−1 ⇒x=((3t−1)/(2−t)) ⇒(dx/dt)=((3(2−t)−(3t−1)(−1))/((2−t)^2 ))=((6−3t+3t−1)/((2−t)^2 ))=(5/((2−t)^2 )) x+3=((3t−1)/(2−t))+3=((3t−1+6−3t)/(2−t))=(5/(2−t)) ⇒ Ψ=∫_(1/3) ^2 (5/((2−t)^2 t^4 ((5/(2−t)))^9 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^7 )/t^4 )dt =(1/5^8 )∫_(1/3) ^2 ((Σ_(k=0) ^7 C_7 ^k (−t)^k 2^(7−k) )/t^4 )dt =(2^7 /5^8 )∫_(1/3) ^2 Σ_(k=0) ^7 C_7 ^k (−1)^k t^(k−4) 2^(−k) dt =(2^7 /5^8 )Σ_(k=0) ^7 (−1)^k 2^(−k) C_7 ^k ∫_(1/3) ^2 t^(k−4) dt =(2^7 /5^8 ){Σ_(k=0 and k≠3) ^7 (−(1/2))^k C_7 ^k [(1/(k−3))t^(k−3) ]_(1/3) ^2 −(1/8)C_7 ^3 [log∣t∣]_(1/3) ^2 } Ψ=(2^7 /5^8 )Σ_(k=0and k≠3) ^7 (−(1/2))^k (C_7 ^k /(k−3))(2^(k−3) −(1/3^(k−3) )) −(2^7 /(8.5^8 ))C_7 ^3 (log2+log3)$
	$Ψ = \int_{0}^{\infty} \frac{dx}{{(2 x + 1)}^{4} {(x + 3)}^{5}} \Rightarrow Ψ = \int_{0}^{\infty} \frac{dx}{{(\frac{2 x + 1}{x + 3})}^{4} {(x + 3)}^{9}}$ $we do the changement \frac{2 x + 1}{x + 3} = t \Rightarrow 2 x + 1 = tx + 3 t \Rightarrow (2 - t) x = 3 t - 1$ $\Rightarrow x = \frac{3 t - 1}{2 - t} \Rightarrow \frac{dx}{dt} = \frac{3 (2 - t) - (3 t - 1) (- 1)}{{(2 - t)}^{2}} = \frac{6 - 3 t + 3 t - 1}{{(2 - t)}^{2}} = \frac{5}{{(2 - t)}^{2}}$ $x + 3 = \frac{3 t - 1}{2 - t} + 3 = \frac{3 t - 1 + 6 - 3 t}{2 - t} = \frac{5}{2 - t} \Rightarrow$ $Ψ = \int_{\frac{1}{3}}^{2} \frac{5}{{(2 - t)}^{2} t^{4} {(\frac{5}{2 - t})}^{9}} dt = \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{{(2 - t)}^{7}}{t^{4}} dt$ $= \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{\sum_{k = 0}^{7} C_{7}^{k} {(- t)}^{k} 2^{7 - k}}{t^{4}} dt$ $= \frac{2^{7}}{5^{8}} \int_{\frac{1}{3}}^{2} \sum_{k = 0}^{7} C_{7}^{k} {(- 1)}^{k} t^{k - 4} 2^{- k} dt$ $= \frac{2^{7}}{5^{8}} \sum_{k = 0}^{7} {(- 1)}^{k} 2^{- k} C_{7}^{k} \int_{\frac{1}{3}}^{2} t^{k - 4} dt$ $= \frac{2^{7}}{5^{8}} {\sum_{k = 0 and k \neq 3}^{7} {(- \frac{1}{2})}^{k} C_{7}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{\frac{1}{3}}^{2} - \frac{1}{8} C_{7}^{3} {[\log ∣ t ∣]}_{\frac{1}{3}}^{2}}$ $Ψ = \frac{2^{7}}{5^{8}} \sum_{k = 0 and k \neq 3}^{7} {(- \frac{1}{2})}^{k} \frac{C_{7}^{k}}{k - 3} (2^{k - 3} - \frac{1}{3^{k - 3}})$ $- \frac{2^{7}}{8 . 5^{8}} C_{7}^{3} (\log 2 + \log 3)$
	Answered by Olaf_Thorendsen last updated on 12/Jul/21

	$R (x) = \frac{1}{{(2 x + 1)}^{4} {(x + 3)}^{5}}$ $R (x) = \frac{\frac{32}{3125}}{{(2 x + 1)}^{4}} - \frac{\frac{32}{3125}}{{(2 x + 1)}^{3}} + \frac{\frac{96}{15625}}{{(2 x + 1)}^{2}}$ $- \frac{\frac{224}{78125}}{2 x + 1} + \frac{\frac{1}{625}}{{(x + 3)}^{5}} + \frac{\frac{8}{3125}}{{(x + 3)}^{4}} + \frac{\frac{8}{3125}}{{(x + 3)}^{3}}$ $+ \frac{\frac{32}{15625}}{{(x + 2)}^{2}} + \frac{\frac{112}{78125}}{x + 3}$ $\int R (x) d x = - \frac{\frac{16}{9375}}{{(2 x + 1)}^{3}} + \frac{\frac{8}{3125}}{{(2 x + 1)}^{2}}$ $- \frac{\frac{48}{15625}}{2 x + 1} - \frac{112}{78125} \ln (2 x + 1) - \frac{\frac{1}{2500}}{{(x + 3)}^{4}}$ $- \frac{\frac{8}{9375}}{{(x + 3)}^{3}} - \frac{\frac{4}{3125}}{{(x + 3)}^{2}} - \frac{\frac{32}{15625}}{x + 3}$ $+ \frac{112}{78125} \ln (x + 3)$ $Ω = \int_{0}^{\infty} R (x) d x = \frac{15593}{5062500} - \frac{112}{78125} \ln 6$
	Commented by mathmax by abdo last updated on 12/Jul/21

	$thank you sir .$
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