1) find ∫ (dx/((x+1)^2 (x−3)^4 )) 2) deduce the decomposition of F(x)=(1/((x+1)^2 (x−3)^4 ))


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Question Number 135957 by mathmax by abdo last updated on 17/Mar/21

$1) find \int \frac{dx}{{(x + 1)}^{2} {(x - 3)}^{4}}$ $2) deduce the decomposition of F (x) = \frac{1}{{(x + 1)}^{2} {(x - 3)}^{4}}$
	Answered by Dwaipayan Shikari last updated on 17/Mar/21

	$\frac{1}{{(x + 1)}^{2} {(x - 3)}^{4}} = \frac{1}{{(x - 3)}^{2}} {(\frac{1}{(x + 1) (x - 3)})}^{2} = τ (x)$ $= \frac{1}{16} (\frac{1}{{(x - 3)}^{2} {(x + 1)}^{2}} - \frac{2}{{(x - 3)}^{3} (x + 1)} + \frac{1}{{(x - 3)}^{4}})$ $= \frac{1}{256} (\frac{1}{{(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{2}} - \frac{2}{(x - 3) (x + 1)}) - \frac{1}{32 {(x - 3)}^{2}} (\frac{1}{x - 3} - \frac{1}{x + 1}) + \frac{1}{16 {(x - 3)}^{4}}$ $= \frac{1}{256 {(x - 3)}^{2}} + \frac{1}{256 {(x + 1)}^{2}} - \frac{1}{512 (x - 3)} + \frac{1}{512 (x + 1)} - \frac{1}{32 {(x - 3)}^{3}} + \frac{1}{128 {(x - 3)}^{2}} - \frac{1}{512 (x - 3)} + \frac{1}{512 (x + 1)} + \frac{1}{16 {(x - 3)}^{4}}$ $= \frac{3}{256 {(x - 3)}^{2}} - \frac{1}{256 (x - 3)} + \frac{1}{256 (x + 1)} + \frac{1}{256 {(x + 1)}^{2}} + \frac{1}{32 {(x - 3)}^{3}} + \frac{1}{16 {(x - 3)}^{4}}$
	Answered by mathmax by abdo last updated on 17/Mar/21
	$1) Φ=∫ (dx/((x+1)^2 (x−3)^4 )) ⇒Φ=∫ (dx/((((x−3)/(x+1)))^4 (x+1)^6 )) we do the changement ((x−3)/(x+1))=t ⇒x−3=tx+t ⇒(1−t)x=3+t ⇒ x=((3+t)/(1−t)) ⇒ (dx/dt)=((1−t−(3+t)(−1))/((1−t)^2 ))=((1−t+3+t)/((1−t)^2 ))=(4/((1−t)^2 )) and x+1 =((3+t)/(1−t))+1 =((3+t+1−t)/(1−t))=(4/(1−t)) ⇒ Φ=∫ (1/(t^4 ((4/(1−t)))^6 ))((4dt)/((1−t)^2 )) =(1/4^5 )∫ (((1−t)^4 )/t^4 )dt =(1/4^5 )∫ (((t−1)^4 )/t^4 ) dt =(1/4^5 )∫ (((t^2 −2t+1)^2 )/t^4 )dt =(1/4^5 )∫ (((t^2 −2t)^2 +2(t^2 −2t)+1)/t^4 )dt =(1/4^5 )∫ ((t^4 −4t^3 +4t^2 +2t^2 −4t +1)/t^4 )dt =(1/4^5 )∫ ((t^4 −4t^3 +6t^2 −4t+1)/t^4 )dt =(1/4^5 ){ ∫ (1−(4/t)+(6/t^2 )−(4/t^3 )+(1/t^4 ))dt} ⇒4^5 .Φ =t−4ln∣t∣−(6/t)−4.(1/(−3+1))t^(−3+1) +(1/(−4+1))t^(−4+1) +C =t−4ln∣t∣−(6/t) +(2/t^2 )−(1/(3t^3 )) +C =((x−3)/(x+1))−4ln∣((x−3)/(x+1))∣ −6.((x+1)/(x−3)) +2(((x+1)/(x−3)))^2 −(1/3)(((x+1)/(x−3)))^3 +C ⇒ Φ=(1/4^5 ){((x−3)/(x+1))−4ln∣((x−3)/(x+1))∣−((6(x+1))/(x−3))+2(((x+1)/(x−3)))^2 −(1/3)(((x+1)/(x−3)))^2 } +C$
	$1) Φ = \int \frac{dx}{{(x + 1)}^{2} {(x - 3)}^{4}} \Rightarrow Φ = \int \frac{dx}{{(\frac{x - 3}{x + 1})}^{4} {(x + 1)}^{6}}$ $we do the changement \frac{x - 3}{x + 1} = t \Rightarrow x - 3 = tx + t \Rightarrow (1 - t) x = 3 + t \Rightarrow$ $x = \frac{3 + t}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{1 - t - (3 + t) (- 1)}{{(1 - t)}^{2}} = \frac{1 - t + 3 + t}{{(1 - t)}^{2}} = \frac{4}{{(1 - t)}^{2}} and$ $x + 1 = \frac{3 + t}{1 - t} + 1 = \frac{3 + t + 1 - t}{1 - t} = \frac{4}{1 - t} \Rightarrow$ $Φ = \int \frac{1}{t^{4} {(\frac{4}{1 - t})}^{6}} \frac{4 dt}{{(1 - t)}^{2}} = \frac{1}{4^{5}} \int \frac{{(1 - t)}^{4}}{t^{4}} dt$ $= \frac{1}{4^{5}} \int \frac{{(t - 1)}^{4}}{t^{4}} dt = \frac{1}{4^{5}} \int \frac{{(t^{2} - 2 t + 1)}^{2}}{t^{4}} dt$ $= \frac{1}{4^{5}} \int \frac{{(t^{2} - 2 t)}^{2} + 2 (t^{2} - 2 t) + 1}{t^{4}} dt$ $= \frac{1}{4^{5}} \int \frac{t^{4} - {4 t}^{3} + {4 t}^{2} + {2 t}^{2} - 4 t + 1}{t^{4}} dt$ $= \frac{1}{4^{5}} \int \frac{t^{4} - {4 t}^{3} + {6 t}^{2} - 4 t + 1}{t^{4}} dt$ $= \frac{1}{4^{5}} {\int (1 - \frac{4}{t} + \frac{6}{t^{2}} - \frac{4}{t^{3}} + \frac{1}{t^{4}}) dt}$ $\Rightarrow 4^{5} . Φ = t - 4 \ln ∣ t ∣ - \frac{6}{t} - 4 . \frac{1}{- 3 + 1} t^{- 3 + 1} + \frac{1}{- 4 + 1} t^{- 4 + 1} + C$ $= t - 4 \ln ∣ t ∣ - \frac{6}{t} + \frac{2}{t^{2}} - \frac{1}{{3 t}^{3}} + C$ $= \frac{x - 3}{x + 1} - 4 \ln ∣ \frac{x - 3}{x + 1} ∣ - 6 . \frac{x + 1}{x - 3} + 2 {(\frac{x + 1}{x - 3})}^{2} - \frac{1}{3} {(\frac{x + 1}{x - 3})}^{3} + C \Rightarrow$ $Φ = \frac{1}{4^{5}} {\frac{x - 3}{x + 1} - 4 \ln ∣ \frac{x - 3}{x + 1} ∣ - \frac{6 (x + 1)}{x - 3} + 2 {(\frac{x + 1}{x - 3})}^{2} - \frac{1}{3} {(\frac{x + 1}{x - 3})}^{2}} + C$
	Commented by mathmax by abdo last updated on 17/Mar/21
	$2) F(x)=(d/dx)Φ (((x−3)/(x+1)))^((1)) =((x+1−(x−3))/((x+1)^2 ))=(4/((x+1)^2 )) =(ln∣((x−3)/(x+1))∣)^((1)) =(4/((x+1)^2 ))×((x+1)/(x−3)) =(4/((x+1)(x−3)))=(1/(x−3))−(1/(x+1)) (((x+1)/(x−3)))^((1)) =((x−3−(x+1))/((x−3)^2 ))=((−4)/((x−3)^2 )) {(((x+1)/(x−3)))^2 }^((1)) =2(((x+1)/(x−3)))×((−4)/((x−3)^2 ))=−((8(x+1))/((x−3)^3 )) =−8((x−3+4)/((x−3)^3 ))=−(8/((x−3)^2 ))−((32)/((x−3)^3 )) (((x+1)/(x−3)))^3 }^((1)) =3(((x+1)/(x−3)))^2 .((−4)/((x−3)^2 )) =−12(((x+1)^2 )/((x−3)^3 )) =−12 (((x−3+4)^2 )/((x−3)^3 )) =−12.(((x−3)^2 +8(x−3)+16)/((x−3)^3 )) =−12{(1/((x−3)))+(8/((x−3)^2 ))+((16)/((x−3)^3 ))} rest to collect the values...$
	$2) F (x) = \frac{d}{dx} Φ$ ${(\frac{x - 3}{x + 1})}^{(1)} = \frac{x + 1 - (x - 3)}{{(x + 1)}^{2}} = \frac{4}{{(x + 1)}^{2}}$ $= {(\ln ∣ \frac{x - 3}{x + 1} ∣)}^{(1)} = \frac{4}{{(x + 1)}^{2}} \times \frac{x + 1}{x - 3} = \frac{4}{(x + 1) (x - 3)} = \frac{1}{x - 3} - \frac{1}{x + 1}$ ${(\frac{x + 1}{x - 3})}^{(1)} = \frac{x - 3 - (x + 1)}{{(x - 3)}^{2}} = \frac{- 4}{{(x - 3)}^{2}}$ ${{(\frac{x + 1}{x - 3})}^{2}}^{(1)} = 2 (\frac{x + 1}{x - 3}) \times \frac{- 4}{{(x - 3)}^{2}} = - \frac{8 (x + 1)}{{(x - 3)}^{3}}$ $= - 8 \frac{x - 3 + 4}{{(x - 3)}^{3}} = - \frac{8}{{(x - 3)}^{2}} - \frac{32}{{(x - 3)}^{3}}$ ${(\frac{x + 1}{x - 3})}^{3}}^{(1)} = 3 {(\frac{x + 1}{x - 3})}^{2} . \frac{- 4}{{(x - 3)}^{2}} = - 12 \frac{{(x + 1)}^{2}}{{(x - 3)}^{3}}$ $= - 12 \frac{{(x - 3 + 4)}^{2}}{{(x - 3)}^{3}} = - 12 . \frac{{(x - 3)}^{2} + 8 (x - 3) + 16}{{(x - 3)}^{3}}$ $= - 12 {\frac{1}{(x - 3)} + \frac{8}{{(x - 3)}^{2}} + \frac{16}{{(x - 3)}^{3}}} rest to collect the values . . .$
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