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


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Question Number 124531 by Bird last updated on 03/Dec/20

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(2 x + 1)}^{4} {(x + 3)}^{5}}$
	Answered by mathmax by abdo last updated on 04/Dec/20
	$let I =∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 )) ⇒I =∫_0 ^∞ (dx/((((2x+1)/(x+3)))^4 (x+3)^9 )) we do the changement ((2x+1)/(x+3))=t ⇒2x+1=tx+3t ⇒(2−t)x=3t−1 ⇒x=((3t−1)/(2−t)) ⇒(dx/dt)=((3(2−t)−(3t−1)(−1))/((2−t)^2 ))=((6−3t+3t−1)/((2−t)^2 ))=(5/((2−t)^2 )) and x+3=((3t−1)/(2−t))+3 =((3t−1+6−3t)/(2−t))=(5/(2−t)) ⇒ I =∫_(1/3) ^2 (5/((2−t)^2 t^4 ((5/(2−t)))^9 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^9 )/(t^4 ((2−t)^2 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^7 )/t^4 )dt =−(1/5^8 )∫_(1/3) ^2 (((t−2)^7 )/t^4 )dt =−(1/5^8 )∫_(1/3) ^2 ((Σ_(k=0) ^7 C_7 ^k t^k (−2)^(7−k) )/t^4 )dt =−(1/5^8 )Σ_(k0) ^7 (−2)^(7−k) C_7 ^k ∫_(1/3) ^2 t^(k−4) dt =−(1/5^8 ) {Σ_(k=0and k≠3) ^7 (−2)^(7−k) C_7 ^k [(1/(k−3))t^(k−3) ]_(1/3) ^2 +(−2)^4 C_7 ^3 [ln2−ln((1/3))) I=−(1/5^8 )Σ_(k=0 and ≠3) ^7 (−2)^(7−k) (C_7 ^k /(k−3)){2^(k−3) −(1/3^(k−3) )} +16 C_7 ^3 ln(6)$
	$let I = \int_{0}^{\infty} \frac{dx}{{(2 x + 1)}^{4} {(x + 3)}^{5}} \Rightarrow I = \int_{0}^{\infty} \frac{dx}{{(\frac{2 x + 1}{x + 3})}^{4} {(x + 3)}^{9}}$ $we do the changement \frac{2 x + 1}{x + 3} = t \Rightarrow 2 x + 1 = tx + 3 t \Rightarrow (2 - t) x = 3 t - 1$ $\Rightarrow x = \frac{3 t - 1}{2 - t} \Rightarrow \frac{dx}{dt} = \frac{3 (2 - t) - (3 t - 1) (- 1)}{{(2 - t)}^{2}} = \frac{6 - 3 t + 3 t - 1}{{(2 - t)}^{2}} = \frac{5}{{(2 - t)}^{2}}$ $and x + 3 = \frac{3 t - 1}{2 - t} + 3 = \frac{3 t - 1 + 6 - 3 t}{2 - t} = \frac{5}{2 - t} \Rightarrow$ $I = \int_{\frac{1}{3}}^{2} \frac{5}{{(2 - t)}^{2} t^{4} {(\frac{5}{2 - t})}^{9}} dt = \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{{(2 - t)}^{9}}{t^{4} ({(2 - t)}^{2}} dt$ $= \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{{(2 - t)}^{7}}{t^{4}} dt = - \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{{(t - 2)}^{7}}{t^{4}} dt$ $= - \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{\sum_{k = 0}^{7} C_{7}^{k} t^{k} {(- 2)}^{7 - k}}{t^{4}} dt$ $= - \frac{1}{5^{8}} \sum_{k 0}^{7} {(- 2)}^{7 - k} C_{7}^{k} \int_{\frac{1}{3}}^{2} t^{k - 4} dt$ $= - \frac{1}{5^{8}} {\sum_{k = 0 and k \neq 3}^{7} {(- 2)}^{7 - k} C_{7}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{\frac{1}{3}}^{2} + {(- 2)}^{4} C_{7}^{3} [\ln 2 - \ln (\frac{1}{3}))$ $I = - \frac{1}{5^{8}} \sum_{k = 0 and \neq 3}^{7} {(- 2)}^{7 - k} \frac{C_{7}^{k}}{k - 3} {2^{k - 3} - \frac{1}{3^{k - 3}}}$ $+ 16 C_{7}^{3} \ln (6)$
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