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Question Number 125593 by shaker last updated on 12/Dec/20

	Answered by MJS_new last updated on 12/Dec/20

	$\int \frac{d x}{{(x^{2} + 2 x + 2)}^{2}} =$ $[{Ostrogradski}^{'} s Method (s e a r c h t h e w w w f o r i t)]$ $= \frac{x + 1}{2 (x^{2} + 2 x + 2)} + \frac{1}{2} \int \frac{d x}{x^{2} + 2 x + 2} =$ $= \frac{x + 1}{2 (x^{2} + 2 x + 2)} + \frac{1}{2} \arctan (x + 1) + C$
	Commented by bemath last updated on 12/Dec/20

	$O s t r o g r a d s k i n i c e m e t h o d$
	Answered by mathmax by abdo last updated on 12/Dec/20
	$parametric method let f(t)=∫ (dx/(x^2 +2x+t)) with ∣t∣>1 we have f^′ (t)=−∫ (dx/((x^2 +2x+t)^2 ))+c ⇒∫ (dx/((x^2 +2x+2)^2 ))=−f^′ (2)+clet explicit f(t) f(t)=∫ (dx/((x+1)^2 +t−1)) =_(x+1=(√(t−1))u) ∫ (((√(t−1))u)/((t−1)(1+u^2 )))du =(1/( (√(t−1))))∫ (du/(1+u^2 )) =((arctan(((x+1)/( (√(t−1))))))/( (√(t−1)))) =((u(t))/(v(t))) f^′ (t)=((u^′ v−uv^′ )/v^2 ) we have u^′ =(((((x+1)/( (√(t−1)))))^′ )/(1+(((x+1)^2 )/(t−1))))=(((x+1)×(−(1/2))(t−1)^(−(3/2)) )/(((x+1)^2 +t−1)/(t−1))) =−((x+1)/2)(t−1)^(−(1/2)) ×(1/((x+1)^2 +t−1)) =−((x+1)/(2(√(t−1)){(x+1)^2 +t−1})) v^′ =(1/(2(√(t−1)))) ⇒f^′ (t)=((−((x+1)/(2((x+1)^2 +t−1)))−(1/(2(√(t−1))))arctan(((x+1)/( (√(t−1))))))/(t−1)) ⇒f^′ (2)=−((x+1)/(2((x+1)^2 +1)))−(1/2)arctan(x+1) +c ⇒ ∫ (dx/((x^2 +2x+2)^2 ))=−f^′ (2)=((x+1)/(2{(x+1)^2 +1}))+(1/2)arctan(x+1) +C$
	$parametric method let f (t) = \int \frac{dx}{x^{2} + 2 x + t} with ∣ t ∣> 1$ $we have f^{^{'}} (t) = - \int \frac{dx}{{(x^{2} + 2 x + t)}^{2}} + c \Rightarrow \int \frac{dx}{{(x^{2} + 2 x + 2)}^{2}} = - f^{^{'}} (2) + clet explicit f (t)$ $f (t) = \int \frac{dx}{{(x + 1)}^{2} + t - 1} =_{x + 1 = \sqrt{t - 1} u} \int \frac{\sqrt{t - 1} u}{(t - 1) (1 + u^{2})} du$ $= \frac{1}{\sqrt{t - 1}} \int \frac{du}{1 + u^{2}} = \frac{\arctan (\frac{x + 1}{\sqrt{t - 1}})}{\sqrt{t - 1}} = \frac{u (t)}{v (t)}$ $f^{^{'}} (t) = \frac{u^{^{'}} v - {uv}^{^{'}}}{v^{2}} we have u^{^{'}} = \frac{{(\frac{x + 1}{\sqrt{t - 1}})}^{^{'}}}{1 + \frac{{(x + 1)}^{2}}{t - 1}} = \frac{(x + 1) \times (- \frac{1}{2}) {(t - 1)}^{- \frac{3}{2}}}{\frac{{(x + 1)}^{2} + t - 1}{t - 1}}$ $= - \frac{x + 1}{2} {(t - 1)}^{- \frac{1}{2}} \times \frac{1}{{(x + 1)}^{2} + t - 1} = - \frac{x + 1}{2 \sqrt{t - 1} {{(x + 1)}^{2} + t - 1}}$ $v^{^{'}} = \frac{1}{2 \sqrt{t - 1}} \Rightarrow f^{^{'}} (t) = \frac{- \frac{x + 1}{2 ({(x + 1)}^{2} + t - 1)} - \frac{1}{2 \sqrt{t - 1}} \arctan (\frac{x + 1}{\sqrt{t - 1}})}{t - 1}$ $\Rightarrow f^{^{'}} (2) = - \frac{x + 1}{2 ({(x + 1)}^{2} + 1)} - \frac{1}{2} \arctan (x + 1) + c \Rightarrow$ $\int \frac{dx}{{(x^{2} + 2 x + 2)}^{2}} = - f^{^{'}} (2) = \frac{x + 1}{2 {{(x + 1)}^{2} + 1}} + \frac{1}{2} \arctan (x + 1) + C$
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