I=∫_0 ^( ∞) (dx/((x+(√(x^2 +1)))^2 )) ?


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Question Number 124579 by bemath last updated on 04/Dec/20

$I = \int_{0}^{\infty} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}} ?$
	Answered by liberty last updated on 04/Dec/20

	$I = \int_{0}^{\infty} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}}$ $l e t t i n g \cot 2 β = x \to d x = - 2 cosec^{2} β d β$ $t h e i n t e g r a l b e c o m e s$ $I = \int_{π / 4}^{0} \frac{- 2 cosec^{2} 2 β d β}{{(\cot 2 β + cosec 2 β)}^{2}}$ $I = \underset{0}{\int^{π / 4}} \frac{1}{{(\cos 2 β + 1)}^{2}} d β = \frac{1}{2} \underset{0}{\int^{π / 4}} \frac{d β}{\cos^{4} β}$ $I = \frac{1}{2} \underset{0}{\int^{π / 4}} \sec^{2} β (\tan^{2} β + 1) d β$ $I = \frac{1}{2} {[\frac{\tan^{3} β}{3} + \tan β]}_{0}^{π / 4} = \frac{2}{3}$
	Commented by peter frank last updated on 05/Dec/20

	$thanks$
	Answered by Bird last updated on 04/Dec/20

	$I = \int_{0}^{\infty} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}} w e d o t h e c h a n g e m e n t$ $x = s h t \Rightarrow I = \int_{0}^{\infty} \frac{c h t}{{(s h t + c h t)}^{2}} d t$ $= \int_{0}^{\infty} \frac{c h t}{{(\frac{e^{t} - e^{- t}}{2} + \frac{e^{t} - e^{- t}}{2})}^{2}} d t$ $= \int_{0}^{\infty} e^{- 2 t} (\frac{e^{t} + e^{- t}}{2}) d t$ $= \frac{1}{2} \int_{0}^{\infty} (e^{- t} + e^{- 3 t}) d t$ $= \frac{1}{2} {[- e^{- t} - \frac{1}{3} e^{- 3 t}]}_{0}^{+ \infty}$ $= \frac{1}{2} (1 + \frac{1}{3}) = \frac{1}{2} (\frac{4}{3}) = \frac{2}{3}$
	Answered by Dwaipayan Shikari last updated on 04/Dec/20

	$A s p e r p r e v i o u s r e s u l t$ $I = {[- \frac{1}{6 {(x + \sqrt{x^{2} + 1})}^{3}} - \frac{1}{2 (x + \sqrt{x^{2} + 1})}]}_{0}^{\infty} = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$
	Answered by Ar Brandon last updated on 04/Dec/20
	$I=∫_0 ^∞ (dx/((x+(√(x^2 +1)))^2 )) t=x+(√(x^2 +1)) , x=((t^2 −1)/(2t)), dx=((t^2 +1)/(2t^2 ))dt I=∫_1 ^∞ (1/t^2 )∙((t^2 +1)/(2t^2 ))dt=(1/2)∫_1 ^∞ {(1/t^2 )+(1/t^4 )}dt =−(1/2){(1/t)+(1/(3t^3 ))}_1 ^∞ =(2/3)$
	$I = \int_{0}^{\infty} \frac{dx}{{(x + \sqrt{x^{2} + 1})}^{2}}$ $t = x + \sqrt{x^{2} + 1}, x = \frac{t^{2} - 1}{2 t}, dx = \frac{t^{2} + 1}{{2 t}^{2}} dt$ $I = \int_{1}^{\infty} \frac{1}{t^{2}} \cdot \frac{t^{2} + 1}{{2 t}^{2}} dt = \frac{1}{2} \int_{1}^{\infty} {\frac{1}{t^{2}} + \frac{1}{t^{4}}} dt$ $= - \frac{1}{2} {\frac{1}{t} + \frac{1}{{3 t}^{3}}}_{1}^{\infty} = \frac{2}{3}$
	Answered by Ar Brandon last updated on 04/Dec/20
	$I=∫_0 ^∞ (dx/((x+(√(x^2 +1)))^2 )) x=tanθ ⇒ dx=sec^2 θdθ I=∫_0 ^(π/2) ((sec^2 θdθ)/((tanθ+(√(tan^2 θ+1)))^2 ))=∫_0 ^(π/2) ((sec^2 θdθ)/((tanθ+secθ)^2 )) =∫_0 ^(π/2) (dθ/((sinθ+1)^2 ))=∫_0 ^(π/2) (((1−sinθ)^2 )/((1−sin^2 θ)^2 ))dθ =∫_0 ^(π/2) {((1−2sinθ+sin^2 θ)/(cos^4 θ))}dθ=∫_0 ^(π/2) {((cos^2 θ−2sinθ+2sin^2 θ)/(cos^4 θ))}dθ =∫_0 ^(π/2) sec^2 θdθ+2∫_0 ^(π/2) (((−sinθ))/(cos^4 θ))dθ+2∫_0 ^(π/2) tan^2 θsec^2 θdθ ={tanθ−(2/(3cos^3 θ))+((2tan^3 θ)/3)}_0 ^(π/2) =(2/3)$
	$I = \int_{0}^{\infty} \frac{dx}{{(x + \sqrt{x^{2} + 1})}^{2}}$ $x = \tan θ \Rightarrow dx = \sec^{2} θ d θ$ $I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} θ d θ}{{(\tan θ + \sqrt{\tan^{2} θ + 1})}^{2}} = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} θ d θ}{{(\tan θ + \sec θ)}^{2}}$ $= \int_{0}^{\frac{π}{2}} \frac{d θ}{{(\sin θ + 1)}^{2}} = \int_{0}^{\frac{π}{2}} \frac{{(1 - \sin θ)}^{2}}{{(1 - \sin^{2} θ)}^{2}} d θ$ $= \int_{0}^{\frac{π}{2}} {\frac{1 - 2 \sin θ + \sin^{2} θ}{\cos^{4} θ}} d θ = \int_{0}^{\frac{π}{2}} {\frac{\cos^{2} θ - 2 \sin θ + {2 \sin}^{2} θ}{\cos^{4} θ}} d θ$ $= \int_{0}^{\frac{π}{2}} \sec^{2} θ d θ + 2 \int_{0}^{\frac{π}{2}} \frac{(- \sin θ)}{\cos^{4} θ} d θ + 2 \int_{0}^{\frac{π}{2}} \tan^{2} θ \sec^{2} θ d θ$ $= {\tan θ - \frac{2}{{3 \cos}^{3} θ} + \frac{{2 \tan}^{3} θ}{3}}_{0}^{\frac{π}{2}} = \frac{2}{3}$
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