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Question Number 99504 by pticantor last updated on 21/Jun/20

	Answered by Ar Brandon last updated on 21/Jun/20
	$Posons x=(√2)tanθ⇒dx=(√2)sec^2 θdθ I=∫(((√2)sec^2 θ)/((2tan^2 θ+2)^2 ))dθ=(1/(2(√2)))∫(1/(sec^2 θ))dθ=(1/(2(√2)))∫cos^2 θdθ =(1/(4(√2)))∫{1+cos(2θ)}dθ=(θ/(4(√2)))+((sin(2θ))/(8(√2)))+C$
	$Posons x = \sqrt{2} \tan θ \Rightarrow dx = \sqrt{2} \sec^{2} θ d θ$ $I = \int \frac{\sqrt{2} \sec^{2} θ}{{({2 \tan}^{2} θ + 2)}^{2}} d θ = \frac{1}{2 \sqrt{2}} \int \frac{1}{\sec^{2} θ} d θ = \frac{1}{2 \sqrt{2}} \int \cos^{2} θ d θ$ $= \frac{1}{4 \sqrt{2}} \int {1 + \cos (2 θ)} d θ = \frac{θ}{4 \sqrt{2}} + \frac{\sin (2 θ)}{8 \sqrt{2}} + C$
	Answered by Dwaipayan Shikari last updated on 21/Jun/20

	$S u p p o s e, x = \sqrt{2} t a n α$ $s o, \frac{d x}{d α} = \sqrt{2} {s e c}^{2} α$ $\int \frac{\sqrt{2} {s e c}^{2} α d α}{4 {({t a n}^{2} α + 1)}^{2}} = \frac{1}{2 \sqrt{2}} \int {c o s}^{2} α d α = \frac{1}{4 \sqrt{2}} \int (1 + c o s 2 α) d α$ $= \frac{α}{4 \sqrt{2}} + \frac{s i n 2 α}{8 \sqrt{2}} + c o n s t a n t$ $= \frac{\tan^{- 1} \frac{x}{\sqrt{2}}}{4 \sqrt{2}} + \frac{s i n ({2 \tan}^{- 1} \frac{x}{\sqrt{2}})}{8 \sqrt{2}} + c o n s t a n t$
	Answered by mathmax by abdo last updated on 21/Jun/20
	$A =∫ (dx/((x^2 +2)^2 )) changement x =(√2)tanθ give A =∫ (((√2)(1+tan^2 θ)dθ)/(4(1+tan^2 θ)^2 )) =((√2)/4) ∫ (dθ/(1+tan^2 θ)) =((√2)/4) ∫cos^2 θ dθ =((√2)/4) ∫((1+cos(2θ))/2)dθ =((√2)/8){ θ +(1/2)sin(2θ)}+c =((√2)/8) arctan((x/(√2))) +((√2)/8)sinθ cosθ+c but sinθ cosθ =tanθ ×cos^2 θ =(x/(√2))((1/(1+tan^2 θ))) =(x/(√2))((1/(1+(x^2 /2)))) =((2x)/((√2)(x^2 +2))) ⇒ A =((√2)/8) arctan((x/(√2)))+((√2)/8)×((2x)/((√2)(x^2 +2)))+c ⇒A =((√2)/8) arctan((x/(√2)))+(x/(4(x^2 +2)))+c$
	$A = \int \frac{dx}{{(x^{2} + 2)}^{2}} changement x = \sqrt{2} \tan θ give A = \int \frac{\sqrt{2} (1 + \tan^{2} θ) d θ}{4 {(1 + \tan^{2} θ)}^{2}}$ $= \frac{\sqrt{2}}{4} \int \frac{d θ}{1 + \tan^{2} θ} = \frac{\sqrt{2}}{4} \int \cos^{2} θ d θ = \frac{\sqrt{2}}{4} \int \frac{1 + \cos (2 θ)}{2} d θ$ $= \frac{\sqrt{2}}{8} {θ + \frac{1}{2} \sin (2 θ)} + c = \frac{\sqrt{2}}{8} \arctan (\frac{x}{\sqrt{2}}) + \frac{\sqrt{2}}{8} \sin θ \cos θ + c but$ $\sin θ \cos θ = \tan θ \times \cos^{2} θ = \frac{x}{\sqrt{2}} (\frac{1}{1 + \tan^{2} θ}) = \frac{x}{\sqrt{2}} (\frac{1}{1 + \frac{x^{2}}{2}}) = \frac{2 x}{\sqrt{2} (x^{2} + 2)} \Rightarrow$ $A = \frac{\sqrt{2}}{8} \arctan (\frac{x}{\sqrt{2}}) + \frac{\sqrt{2}}{8} \times \frac{2 x}{\sqrt{2} (x^{2} + 2)} + c \Rightarrow A = \frac{\sqrt{2}}{8} \arctan (\frac{x}{\sqrt{2}}) + \frac{x}{4 (x^{2} + 2)} + c$
	Answered by maths mind last updated on 22/Jun/20

	$f (t) = \int \frac{d x}{x^{2} + t^{2}} = \frac{1}{t} \tan^{- 1} (\frac{x}{t}) + c$ $f^{'} (t) = \int \frac{- 2 t d x}{{(x^{2} + t^{2})}^{2}} = - \frac{1}{t^{2}} {t a n}^{-} (\frac{x}{t}) - \frac{x}{t} . \frac{1}{x^{2} + t^{2}} + c$ $t = \sqrt{2} \Rightarrow \int \frac{d x}{{(x^{2} + 2)}^{2}} = \frac{1}{4 \sqrt{2}} {t a n}^{-} (\frac{x}{\sqrt{2}}) + \frac{x}{4 (x^{2} + 2)} + d$
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