1)decompose F(x)=(1/((x+1)^5 (2x−1)^4 )) 2) find ∫


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Question Number 133957 by mathmax by abdo last updated on 25/Feb/21

$1) decompose F (x) = \frac{1}{{(x + 1)}^{5} {(2 x - 1)}^{4}}$ $2) find \int_{1}^{\infty} F (x) dx$
	Answered by Olaf last updated on 25/Feb/21

	$F (x) = \frac{1}{{(x + 1)}^{5} {(2 x - 1)}^{4}}$ $F (x) = \frac{1}{81 {(x + 1)}^{5}} + \frac{8}{243 {(x + 1)}^{4}} + \frac{40}{729 {(x + 1)}^{3}}$ $+ \frac{160}{2187 {(x + 1)}^{2}} + \frac{570}{6561 (x + 1)} + \frac{2}{243 {(x - \frac{1}{2})}^{4}}$ $- \frac{20}{729 {(x - \frac{1}{2})}^{3}} + \frac{40}{729 {(x - \frac{1}{2})}^{2}} - \frac{560}{6561 (x - \frac{1}{2})}$ $\int_{1}^{\infty} F (x) d x = [- \frac{1}{324 {(x + 1)}^{4}} - \frac{8}{729 {(x + 1)}^{3}}$ $- \frac{20}{729 {(x + 1)}^{2}} - \frac{160}{2187 (x + 1)} + \frac{570}{6561} \ln (x + 1)$ $- \frac{2}{729 {(x - \frac{1}{2})}^{3}} + \frac{10}{729 {(x - \frac{1}{2})}^{2}} - \frac{40}{729 (x - \frac{1}{2})}$ ${- \frac{560}{6561} \ln (x - \frac{1}{2})]}_{1}^{\infty}$ $= \frac{27051}{139968} - \frac{1120}{6561} \ln 2$
	Commented by mathmax by abdo last updated on 26/Feb/21

	$thank you sir$
	Answered by mathmax by abdo last updated on 27/Feb/21
	$2)Φ=∫_1 ^∞ (dx/((x+1)^5 (2x−1)^4 )) ⇒Φ =∫_1 ^∞ (dx/((((x+1)/(2x−1)))^5 (2x−1)^9 )) we do the changement ((x+1)/(2x−1))=t ⇒x+1=2tx−t ⇒(1−2t)x=−1−t ⇒ x=((−1−t)/(1−2t)) =((t+1)/(2t−1)) ⇒(dx/dt)=((2t−1−2(t+1))/((2t−1)^2 ))=((−3)/((2t−1)^2 )) and 2x−1 =((2t+2)/(2t−1))−1 =((2t+2−2t+1)/(2t−1)) =(3/(2t−1)) ⇒ Φ =∫_2 ^(1/2) (1/(t^5 ((3/(2t−1)))^9 ))((−3dt)/((2t−1)^2 )) =(1/3^8 )∫_(1/2) ^2 (((2t−1)^9 )/((2t−1)^2 t^5 ))dt =(1/3^8 )∫_(1/2) ^2 (((2t−1)^7 )/t^5 )dt ⇒3^8 Φ =∫_(1/2) ^2 (1/t^5 )Σ_(k=0) ^7 C_7 ^k (2t)^k (−1)^(7−k) =−Σ_(k=0) ^7 2^k (−1)^k C_7 ^k ∫_(1/2) ^2 t^(k−5) dt =−Σ_(k=0 and k≠4) ^7 (−2)^k C_7 ^k [(1/(k−4))t^(k−4) ]_(1/2) ^2 −(−2)^4 C_7 ^4 [ln∣t∣]_(1/2) ^2 =−Σ_(k=0 and k≠4) ^7 (−2)^k (C_7 ^k /(k−4))(2^(k−4) −(1/2^(k−4) ))−2^4 C_7 ^4 (2ln2) ⇒ Φ =−(1/3^8 ){ Σ_(k=0 and k≠4) ^7 (((−2)^k C_7 ^k )/(k−4))(2^(k−4) −(1/2^(k−4) ))+2^5 ln(2)C_7 ^4 }$
	$2) Φ = \int_{1}^{\infty} \frac{dx}{{(x + 1)}^{5} {(2 x - 1)}^{4}} \Rightarrow Φ = \int_{1}^{\infty} \frac{dx}{{(\frac{x + 1}{2 x - 1})}^{5} {(2 x - 1)}^{9}}$ $we do the changement \frac{x + 1}{2 x - 1} = t \Rightarrow x + 1 = 2 tx - t \Rightarrow (1 - 2 t) x = - 1 - t \Rightarrow$ $x = \frac{- 1 - t}{1 - 2 t} = \frac{t + 1}{2 t - 1} \Rightarrow \frac{dx}{dt} = \frac{2 t - 1 - 2 (t + 1)}{{(2 t - 1)}^{2}} = \frac{- 3}{{(2 t - 1)}^{2}} and$ $2 x - 1 = \frac{2 t + 2}{2 t - 1} - 1 = \frac{2 t + 2 - 2 t + 1}{2 t - 1} = \frac{3}{2 t - 1} \Rightarrow$ $Φ = \int_{2}^{\frac{1}{2}} \frac{1}{t^{5} {(\frac{3}{2 t - 1})}^{9}} \frac{- 3 dt}{{(2 t - 1)}^{2}} = \frac{1}{3^{8}} \int_{\frac{1}{2}}^{2} \frac{{(2 t - 1)}^{9}}{{(2 t - 1)}^{2} t^{5}} dt$ $= \frac{1}{3^{8}} \int_{\frac{1}{2}}^{2} \frac{{(2 t - 1)}^{7}}{t^{5}} dt \Rightarrow 3^{8} Φ = \int_{\frac{1}{2}}^{2} \frac{1}{t^{5}} \sum_{k = 0}^{7} C_{7}^{k} {(2 t)}^{k} {(- 1)}^{7 - k}$ $= - \sum_{k = 0}^{7} 2^{k} {(- 1)}^{k} C_{7}^{k} \int_{\frac{1}{2}}^{2} t^{k - 5} dt$ $= - \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} C_{7}^{k} {[\frac{1}{k - 4} t^{k - 4}]}_{\frac{1}{2}}^{2} - {(- 2)}^{4} C_{7}^{4} {[\ln ∣ t ∣]}_{\frac{1}{2}}^{2}$ $= - \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} \frac{C_{7}^{k}}{k - 4} (2^{k - 4} - \frac{1}{2^{k - 4}}) - 2^{4} C_{7}^{4} (2 \ln 2) \Rightarrow$ $Φ = - \frac{1}{3^{8}} {\sum_{k = 0 and k \neq 4}^{7} \frac{{(- 2)}^{k} C_{7}^{k}}{k - 4} (2^{k - 4} - \frac{1}{2^{k - 4}}) + 2^{5} \ln (2) C_{7}^{4}}$
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