Calculate ; J=∫(dx/(x(x^2 +x−1)^2 )) K=∫((x^3 +x−1)/((x^2 +2)^2 ))dx L=∫(dx/(x+(√(x^2 +1))))


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Question Number 102462 by Ar Brandon last updated on 09/Jul/20

$Calculate;$ $J = \int \frac{dx}{x {(x^{2} + x - 1)}^{2}}$ $K = \int \frac{x^{3} + x - 1}{{(x^{2} + 2)}^{2}} dx$ $L = \int \frac{dx}{x + \sqrt{x^{2} + 1}}$
	Answered by bemath last updated on 09/Jul/20
	$L=((x−(√(x^2 +1)))/(x^2 −(x^2 +1)))= (((√(x^2 +1))−x)/1) ∫ {(√(x^2 +1))−x} dx = I_1 = ∫(√(x^2 +1)) dx x= tan p ⇒dx = sec^2 p dp I_1 = ∫sec^3 p dp = ((sin p)/(2cos^2 p))+(1/2)ln∣tan ((p/2)+(π/4))∣ I_2 = ∫x dx = (1/2)x^2 I=I_1 −I_2$
	$L = \frac{x - \sqrt{x^{2} + 1}}{x^{2} - (x^{2} + 1)} = \frac{\sqrt{x^{2} + 1} - x}{1}$ $\int {\sqrt{x^{2} + 1} - x} d x =$ $I_{1} = \int \sqrt{x^{2} + 1} d x$ $x = \tan p \Rightarrow d x = \sec^{2} p d p$ $I_{1} = \int \sec^{3} p d p = \frac{\sin p}{2 \cos^{2} p} + \frac{1}{2} \ln ∣ \tan (\frac{p}{2} + \frac{π}{4}) ∣$ $I_{2} = \int x d x = \frac{1}{2} x^{2}$ $I = I_{1} - I_{2}$
	Answered by Dwaipayan Shikari last updated on 09/Jul/20

	$L = \int \frac{x - \sqrt{x^{2} + 1}}{x^{2} - x^{2} - 1} = \int \sqrt{x^{2} + 1} d x - \int x d x = \frac{x}{2} \sqrt{x^{2} + 1} + \frac{1}{2} l o g (x + \sqrt{x^{2} + 1}) - \frac{x^{2}}{2} + C o n s t a n t$
	Answered by PRITHWISH SEN 2 last updated on 09/Jul/20

	$J = \int \frac{dx}{x} - \int \frac{(x + 1)}{x^{2} + x - 1} dx + \int \frac{(x + 1)}{{(x^{2} + x - 1)}^{2}} dx$ $For K the decomposition leads to$ $K = \frac{1}{4} \int \frac{{4 x}^{3} + 8 x}{{(x^{2} + 2)}^{2}} dx - \frac{1}{2} \int \frac{2 x}{{(x^{2} + 2)}^{2}} - \int \frac{dx}{{(x^{2} + 2)}^{2}}$ $= \frac{1}{4} \ln ∣ x^{4} + 4 x^{2} + 4 ∣ + \frac{1}{2 (x^{2} + 2)} - \frac{x}{4 (x^{2} + 2)} - \frac{1}{4 \sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) + C$
	Commented by Ar Brandon last updated on 09/Jul/20
	Thanks for your time.��
	Answered by 1549442205 last updated on 09/Jul/20

	$K = \int \frac{x (x^{2} + 2) - (x + 1)}{{(x^{2} + 2)}^{2}} dx = \int \frac{x}{x^{2} + 2} dx - \int \frac{x + 1}{{(x^{2} + 2)}^{2}} dx$ $Missing \left or extra \right$ $K = \frac{1}{2} \ln (x^{2} + 2) + \frac{1}{2 (x^{2} + 2)} - \frac{x}{4 (x^{2} + 2)} - \frac{1}{4} \times \frac{1}{\sqrt{2}} \arctan \frac{x}{\sqrt{2}}$ $= \frac{1}{2} \ln (x^{2} + 2) + \frac{2 - x}{4 (x^{2} + 2)} - \frac{1}{4 \sqrt{2}} \arctan \frac{x}{\sqrt{2}} + C$
	Answered by Dwaipayan Shikari last updated on 09/Jul/20
	$K=∫((x^3 +x)/(x^4 +2x^2 +4))dx−{∫(1/((x^2 +2)^2 ))dx}=I_a K=(1/4)∫((4x^3 +4x)/(x^4 +2x^2 +4))−I_(a ) (√2)tanθ=x K=−(1/4).(1/((x^2 +2)))−I_a {I_a =∫(1/((x^2 +2)^2 ))=(1/4)∫(((√2)sec^2 θdθ)/((tan^2 θ+1)^2 )) K=−(1/4).(1/((x^2 +2)))−(1/(4(√2)))tan^(−1) (x/(√2))−(1/(8(√2)))sin(2tan^(−1) (x/(√2)))+C {I_a =(1/(2(√2)))∫cos^2 θdθ=(1/(2(√2)))((θ/2)+(1/4)sin2θ)$
	$K = \int \frac{x^{3} + x}{x^{4} + 2 x^{2} + 4} d x - {\int \frac{1}{{(x^{2} + 2)}^{2}} d x} = I_{a}$ $K = \frac{1}{4} \int \frac{4 x^{3} + 4 x}{x^{4} + 2 x^{2} + 4} - I_{a} \sqrt{2} t a n θ = x$ $K = - \frac{1}{4} . \frac{1}{(x^{2} + 2)} - I_{a} {I_{a} = \int \frac{1}{{(x^{2} + 2)}^{2}} = \frac{1}{4} \int \frac{\sqrt{2} {s e c}^{2} θ d θ}{{({t a n}^{2} θ + 1)}^{2}}$ $K = - \frac{1}{4} . \frac{1}{(x^{2} + 2)} - \frac{1}{4 \sqrt{2}} {t a n}^{- 1} \frac{x}{\sqrt{2}} - \frac{1}{8 \sqrt{2}} s i n (2 {t a n}^{- 1} \frac{x}{\sqrt{2}}) + C {I_{a} = \frac{1}{2 \sqrt{2}} \int {c o s}^{2} θ d θ = \frac{1}{2 \sqrt{2}} (\frac{θ}{2} + \frac{1}{4} s i n 2 θ)$
	Answered by mathmax by abdo last updated on 09/Jul/20

	$let f (a) = \int \frac{dx}{x (x^{2} + x + a)} with a < \frac{1}{4} f^{^{'}} (a) = - \int \frac{dx}{x {(x^{2} + x + a)}^{2}}$ $and J = - f^{^{'}} (- 1) let explicit f (a) first we decompose F (x) = \frac{1}{x (x^{2} + x + a)}$ $x^{2} + x + a = 0 \to Δ = 1 - 4 a > 0 \Rightarrow x_{1} = \frac{- 1 + \sqrt{1 - 4 a}}{2}$ $x_{2} = \frac{- 1 - \sqrt{1 - 4 a}}{2} \Rightarrow F (x) = \frac{1}{x (x - x_{1}) (x - x_{2})} = \frac{α}{x} + \frac{β}{x - x_{1}} + \frac{λ}{x - x_{2}}$ $α = \frac{1}{a}, β = \frac{1}{x_{1} \sqrt{1 - 4 a}} = \frac{2}{(\sqrt{1 - 4 a} - 1) \sqrt{1 - 4 a})}$ $λ = \frac{1}{x_{2} (x_{2} - x_{1})} = \frac{2}{(- 1 - \sqrt{1 - 4 a}) (- \sqrt{1 - 4 a})} = \frac{2}{(1 + \sqrt{1 - 4 a}) \sqrt{1 - 4 a}} \Rightarrow$ $f (a) = α \ln ∣ x ∣ + β \ln ∣ x - x_{1} ∣ + λ \ln ∣ x - x_{2} ∣ + c$ $\Rightarrow f (a) = \frac{1}{a} \ln ∣ x ∣ + \frac{2}{(\sqrt{1 - 4 a} - 1) \sqrt{1 - 4 a}} \ln ∣ x - \frac{- 1 + \sqrt{1 - 4 a}}{2} ∣$ $+ \frac{2}{(1 + \sqrt{1 - 4 a}) \sqrt{1 - 4 a}} \ln ∣ x + \frac{1 + \sqrt{1 - 4 a}}{2} ∣ + c rest to calculate f^{^{'}} (a)$ $. . . be continued . . .$
	Answered by mathmax by abdo last updated on 09/Jul/20

	$L = \int \frac{dx}{x + \sqrt{1 + x^{2}}} we do the cha 7 gement x = sht \Rightarrow$ $L = \int \frac{ch (t)}{sht + ch (t)} dt = \int \frac{\frac{e^{t} + e^{- t}}{2}}{\frac{e^{t} - e^{- t}}{2} + \frac{e^{t} + e^{- t}}{2}} dt$ $= \int \frac{e^{t} + e^{- t}}{e^{t}} dt = \int (1 + e^{- 2 t}) dt = t - \frac{1}{2} e^{- 2 t} + c$ $we have t = argsh (x) = \ln (x + \sqrt{1 + x^{2}}) \Rightarrow e^{2 t} = {(x + \sqrt{1 + x^{2}})}^{2} \Rightarrow$ $L = \ln (x + \sqrt{1 + x^{2}}) - \frac{1}{2 {(x + \sqrt{1 + x^{2}})}^{2}} + C$
	Answered by mathmax by abdo last updated on 13/Jul/20
	$snother way for J =∫ (dx/(x(x^2 +x−1)^2 )) x^2 +x−1 =0→Δ =1+4 =5 ⇒α =((−1+(√5))/2) and β =((−1−(√5))/2) ⇒J =∫ (dx/(x(x−α)^2 (x−β)^2 )) =∫ (dx/(x(((x−α)/(x−β)))^2 (x−β)^4 )) we do the changement ((x−α)/(x−β))=t ⇒x−α =tx−βt ⇒(1−t)x =−βt +α ⇒x =((βt−α)/(t−1)) and x−β =((βt−α)/(t−1))−β =((βt−α−βt+β)/(t−1)) =((β−α)/(t−1)) ⇒ (dx/dt) =((β(t−1)−βt +α)/((t−1)^2 )) =((α−β)/((t−1)^2 )) ⇒ J =∫ ((α−β)/((t−1)^2 t^2 (((β−α)/(t−1)))^4 )) dt =(1/((α−β)^3 )) ∫ (((t−1)^4 )/((t−1)^2 t^2 ))dt =(1/((α−β)^3 )) ∫ (((t−1)^2 )/t^2 ) dt =(1/((α−β)^3 )) ∫ (((t^2 −2t +1)/t^2 ))dt =(1/((α−β)^3 )) ∫( 1−(2/t) +(1/t^2 ))dt =(1/((α−β)^3 )) { t−2ln∣t∣−(1/t)} +c =(1/(((√5))^3 )){ ((x−((−1+(√5))/2))/(x+((1+(√5))/2))) −2ln∣((x−((−1+(√5))/2))/(x+((1+(√5))/2)))∣−((x+((1+(√5))/2))/(x−((−1+(√5))/2)))} +c J=(1/(5(√5))){ ((2x+1−(√5))/(2x+1+(√5))) −2ln∣((2x+1−(√5))/(2x+1+(√5)))∣−((2x+1+(√5))/(2x+1−(√5)))} +c$
	$snother way for J = \int \frac{dx}{x {(x^{2} + x - 1)}^{2}}$ $x^{2} + x - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow α = \frac{- 1 + \sqrt{5}}{2} and β = \frac{- 1 - \sqrt{5}}{2}$ $\Rightarrow J = \int \frac{dx}{x {(x - α)}^{2} {(x - β)}^{2}} = \int \frac{dx}{x {(\frac{x - α}{x - β})}^{2} {(x - β)}^{4}} we do the changement$ $\frac{x - α}{x - β} = t \Rightarrow x - α = tx - β t \Rightarrow (1 - t) x = - β t + α \Rightarrow x = \frac{β t - α}{t - 1}$ $and x - β = \frac{β t - α}{t - 1} - β = \frac{β t - α - β t + β}{t - 1} = \frac{β - α}{t - 1} \Rightarrow$ $\frac{dx}{dt} = \frac{β (t - 1) - β t + α}{{(t - 1)}^{2}} = \frac{α - β}{{(t - 1)}^{2}} \Rightarrow$ $J = \int \frac{α - β}{{(t - 1)}^{2} t^{2} {(\frac{β - α}{t - 1})}^{4}} dt = \frac{1}{{(α - β)}^{3}} \int \frac{{(t - 1)}^{4}}{{(t - 1)}^{2} t^{2}} dt$ $= \frac{1}{{(α - β)}^{3}} \int \frac{{(t - 1)}^{2}}{t^{2}} dt = \frac{1}{{(α - β)}^{3}} \int (\frac{t^{2} - 2 t + 1}{t^{2}}) dt$ $= \frac{1}{{(α - β)}^{3}} \int (1 - \frac{2}{t} + \frac{1}{t^{2}}) dt = \frac{1}{{(α - β)}^{3}} {t - 2 \ln ∣ t ∣ - \frac{1}{t}} + c$ $= \frac{1}{{(\sqrt{5})}^{3}} {\frac{x - \frac{- 1 + \sqrt{5}}{2}}{x + \frac{1 + \sqrt{5}}{2}} - 2 \ln ∣ \frac{x - \frac{- 1 + \sqrt{5}}{2}}{x + \frac{1 + \sqrt{5}}{2}} ∣ - \frac{x + \frac{1 + \sqrt{5}}{2}}{x - \frac{- 1 + \sqrt{5}}{2}}} + c$ $J = \frac{1}{5 \sqrt{5}} {\frac{2 x + 1 - \sqrt{5}}{2 x + 1 + \sqrt{5}} - 2 \ln ∣ \frac{2 x + 1 - \sqrt{5}}{2 x + 1 + \sqrt{5}} ∣ - \frac{2 x + 1 + \sqrt{5}}{2 x + 1 - \sqrt{5}}} + c$
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