calculate ∫_1 ^(+∞) (dx/(x^3 (2x+1)^4 )) 1)without use of decomposition 2)by use of decomposition


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Question Number 95584 by turbo msup by abdo last updated on 26/May/20

$c a l c u l a t e \int_{1}^{+ \infty} \frac{d x}{x^{3} {(2 x + 1)}^{4}}$ $1) w i t h o u t u s e o f d e c o m p o s i t i o n$ $2) b y u s e o f d e c o m p o s i t i o n$
	Answered by MJS last updated on 26/May/20

	$Ostrogradski leads to$ $\frac{960 x^{4} + 1200 x^{3} + 440 x^{2} + 30 x - 3}{6 x^{2} {(2 x + 1)}^{3}} + 40 \int \frac{d x}{x (2 x + 1)} =$ $= \frac{960 x^{4} + 1200 x^{3} + 440 x^{2} + 30 x - 3}{6 x^{2} {(2 x + 1)}^{3}} + 40 \ln \frac{x}{2 x + 1} + C$ $\Rightarrow$ $\underset{1}{\int^{+ \infty}} \frac{d x}{x^{3} {(2 x + 1)}^{4}} = 40 \ln \frac{3}{2} - \frac{2627}{162}$
	Commented by mathmax by abdo last updated on 26/May/20

	$thank you sir mjs$
	Answered by mathmax by abdo last updated on 26/May/20
	$A =∫_1 ^(+∞) (dx/(x^3 (2x+1)^4 ))⇒A =∫_1 ^(+∞) (dx/(((x/(2x+1)))^3 (2x+1)^7 )) we do the changement (x/(2x+1)) =t ⇒ x =2tx +t ⇒(1−2t)x =t ⇒x =(t/(1−2t)) ⇒(dx/dt) =((1−2t −t(−2t))/((1−2t)^2 )) =(1/((1−2t)^2 )) and 2x+1 =((2t)/(1−2t)) +1 =(1/(1−2t)) ⇒ A = ∫_(1/3) ^(1/2) (dt/((2t−1)^2 t^3 ((1/(1−2t)))^7 )) =−∫_(1/3) ^(1/2) (((2t−1)^7 )/((2t−1)^2 t^3 ))dt =−∫_(1/3) ^(1/2) (((2t−1)^5 )/t^3 ) dt =−∫_(1/3) ^(1/2) ((Σ_(k=0) ^5 C_5 ^k (2t)^k (−1)^(5−k) )/t^3 )dt =∫_(1/3) ^(1/2) ((Σ_(k=0) ^5 2^k (−1)^k C_5 ^k t^k )/t^3 )dt =Σ_(k=0) ^5 C_5 ^k (−2)^k ∫_(1/3) ^(1/2) t^(k−3) dt =Σ_(k=0 and k≠2) ^5 (−2)^k C_5 ^k [(1/(k−2))t^(k−2) ]_(1/3) ^(1/2) +4 C_5 ^2 [lnt]_(1/3) ^(1/2) =Σ_(k=0 and k≠2) ^5 (−2)^k (C_5 ^k /(k−2)){ ((1/2))^(k−2) −((1/3))^(k−2) }+4C_5 ^2 {ln((1/2))−ln((1/3))}$
	$A = \int_{1}^{+ \infty} \frac{dx}{x^{3} {(2 x + 1)}^{4}} \Rightarrow A = \int_{1}^{+ \infty} \frac{dx}{{(\frac{x}{2 x + 1})}^{3} {(2 x + 1)}^{7}} we do the changement \frac{x}{2 x + 1} = t \Rightarrow$ $x = 2 tx + t \Rightarrow (1 - 2 t) x = t \Rightarrow x = \frac{t}{1 - 2 t} \Rightarrow \frac{dx}{dt} = \frac{1 - 2 t - t (- 2 t)}{{(1 - 2 t)}^{2}}$ $= \frac{1}{{(1 - 2 t)}^{2}} and 2 x + 1 = \frac{2 t}{1 - 2 t} + 1 = \frac{1}{1 - 2 t} \Rightarrow$ $A = \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{dt}{{(2 t - 1)}^{2} t^{3} {(\frac{1}{1 - 2 t})}^{7}} = - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{{(2 t - 1)}^{7}}{{(2 t - 1)}^{2} t^{3}} dt$ $= - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{{(2 t - 1)}^{5}}{t^{3}} dt = - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\sum_{k = 0}^{5} C_{5}^{k} {(2 t)}^{k} {(- 1)}^{5 - k}}{t^{3}} dt$ $= \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\sum_{k = 0}^{5} 2^{k} {(- 1)}^{k} C_{5}^{k} t^{k}}{t^{3}} dt$ $= \sum_{k = 0}^{5} C_{5}^{k} {(- 2)}^{k} \int_{\frac{1}{3}}^{\frac{1}{2}} t^{k - 3} dt$ $= \sum_{k = 0 and k \neq 2}^{5} {(- 2)}^{k} C_{5}^{k} {[\frac{1}{k - 2} t^{k - 2}]}_{\frac{1}{3}}^{\frac{1}{2}} + 4 C_{5}^{2} {[lnt]}_{\frac{1}{3}}^{\frac{1}{2}}$ $= \sum_{k = 0 and k \neq 2}^{5} {(- 2)}^{k} \frac{C_{5}^{k}}{k - 2} {{(\frac{1}{2})}^{k - 2} - {(\frac{1}{3})}^{k - 2}} + {4 C}_{5}^{2} {\ln (\frac{1}{2}) - \ln (\frac{1}{3})}$
	Answered by mathmax by abdo last updated on 27/May/20

	$thank you sir mjs$ $2) let decompose F (x) = \frac{1}{x^{3} {(2 x + 1)}^{4}} \Rightarrow$ $2^{4} F (x) = \frac{1}{x^{3} {(x + \frac{1}{2})}^{4}} = \sum_{i = 1}^{3} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{4} \frac{b_{i}}{{(x + \frac{1}{2})}^{i}}$ $a_{i} ? we find D_{2} (0) for f (x) = {(x + \frac{1}{2})}^{- 4}$ $f (x) = f (0) + \frac{x}{1!} f^{^{'}} (0) + \frac{x^{2}}{2!} f^{(2)} (0) + \frac{x^{3}}{3!} ξ (x)$ $f (0) = \frac{1}{2^{- 4}}, f^{^{'}} (x) = - 4 {(x + \frac{1}{2})}^{- 5} \Rightarrow f^{^{'}} (0) = \frac{- 4}{2^{- 5}}$ $f^{(2)} (x) = 20 {(x + \frac{1}{2})}^{- 6} \Rightarrow f^{(2)} (0) = \frac{20}{2^{- 6}} \Rightarrow$ $f (x) = 2^{4} - 4 . 2^{5} x + 10 . 2^{6} x^{2} + \frac{x^{3}}{3!} ξ (x) \Rightarrow$ $\frac{f (x)}{x^{3}} = \frac{2^{4}}{x^{3}} - \frac{4 . 2^{5}}{x^{2}} + \frac{10 . 2^{6}}{x} + \frac{1}{3!} ξ (x) \Rightarrow a_{1} = 10 . 2^{6}$ $a_{2} = - 4 . 2^{5} and a_{3} = 2^{4} let find b_{i}$ $we determine D_{3} (- \frac{1}{2}) for g (x) = x^{- 3}$ $g (x) = g (- \frac{1}{2}) + \frac{x + \frac{1}{2}}{1!} g^{^{'}} (- \frac{1}{2}) + \frac{{(x + \frac{1}{2})}^{2}}{2!} g^{(2)} (- \frac{1}{2})$ $+ \frac{{(x + \frac{1}{2})}^{3}}{3!} g^{(3)} (- \frac{1}{2}) + \frac{{(x + \frac{1}{2})}^{4}}{4!} ξ (x)$ $g (- \frac{1}{2}) = {(- \frac{1}{2})}^{- 3}, g^{^{'}} (x) = - {3 x}^{- 4} \Rightarrow g^{^{'}} (- \frac{1}{2}) = - 3 {(- \frac{1}{2})}^{- 4}$ $g^{(2)} (x) = 12 x^{- 5} \Rightarrow g^{(2)} (- \frac{1}{2}) = 12 {(- \frac{1}{2})}^{- 5}$ $g^{(3)} (x) = - 60 x^{- 6} \Rightarrow g^{(3)} (- \frac{1}{2}) = - 60 {(- \frac{1}{2})}^{- 6}$ $\Rightarrow g (x) = {(- \frac{1}{2})}^{- 3} - 3 {(- \frac{1}{2})}^{- 4} (x + \frac{1}{2}) + 6 {(- \frac{1}{2})}^{- 5} {(x + \frac{1}{2})}^{2}$ $- 10 {(- \frac{1}{2})}^{- 6} {(x + \frac{1}{2})}^{3} + \frac{{(x + \frac{1}{2})}^{4}}{4!} ξ (x)$ $= - 2^{3} - 3 . 2^{4} (x + \frac{1}{2}) - 6 . 2^{5} {(x + \frac{1}{2})}^{2} - 10 . 2^{6} {(x + \frac{1}{2})}^{3}$ $+ \frac{{(x + \frac{1}{2})}^{4}}{4!} ξ (x) \Rightarrow$ $\frac{g (x)}{{(x + \frac{1}{2})}^{4}} = - \frac{2^{3}}{{(x + \frac{1}{2})}^{4}} - \frac{3 . 2^{4}}{{(x + \frac{1}{2})}^{3}} - \frac{6 . 2^{5}}{{(x + \frac{1}{2})}^{2}} - \frac{10 . 2^{6}}{(x + \frac{1}{2})} + . . \Rightarrow$ $\Rightarrow b_{1} = - 10 . 2^{6}, b_{2} = - 6 . 2^{5}, b_{3} = - 3 . 2^{4}, b_{4} = - 2^{3} \Rightarrow$ $2^{4} F (x) = \frac{10 . 2^{6}}{x} - \frac{4 . 2^{5}}{x^{2}} + \frac{2^{4}}{x^{3}} - \frac{10 . 2^{6}}{(x + \frac{1}{2})} - \frac{6 . 2^{5}}{{(x + \frac{1}{2})}^{2}} - \frac{3 . 2^{4}}{{(x + \frac{1}{2})}^{3}}$ $- \frac{2^{3}}{{(x + \frac{1}{2})}^{4}} \Rightarrow$ $F (x) = \frac{10 . 2^{2}}{x} - \frac{4.2}{x^{2}} + \frac{1}{x^{3}} - \frac{10 . 2^{2}}{(x + \frac{1}{2})} - \frac{6.2}{{(x + \frac{1}{2})}^{2}} - \frac{3}{{(x + \frac{1}{2})}^{3}}$ $- \frac{2^{- 1}}{{(x + \frac{1}{2})}^{4}} now its eazy to find \int_{1}^{+ \infty} F (x) dx . . .$
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