∫ (dx/((x+(√(x^2 +1)))^2 ))


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Question Number 124566 by bemath last updated on 04/Dec/20

$\int \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}}$
	Answered by liberty last updated on 05/Dec/20

	$\int \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}} = ?$ $\frac{1}{x + \sqrt{x^{2} + 1}} = \frac{x - \sqrt{x^{2} + 1}}{x^{2} - (x^{2} + 1)} = \sqrt{x^{2} + 1} - x$ $\int \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}} = \int {(\sqrt{x^{2} + 1} - x)}^{2} d x$ $\int 2 x^{2} + 1 - 2 x \sqrt{x^{2} + 1} d x = \frac{2}{3} x^{3} + x - \frac{2}{3} (x^{2} + 1) \sqrt{x^{2} + 1} + c$ $= \frac{2}{3} (x^{2} + 1) [x - \sqrt{x^{2} + 1}] + c$ $= \frac{2}{3} (x^{2} + 1) [\frac{- 1}{x + \sqrt{x^{2} + 1}}] + c$ $= - \frac{2 (x^{2} + 1)}{3 (x + \sqrt{x^{2} + 1})} + c$ $I f \underset{0}{\int^{\infty}} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{2}} = \lim_{p \to \infty} - \frac{2}{3} {[\frac{x^{2} + 1}{x + \sqrt{x^{2} + 1}}]}_{0}^{p}$ $= - \frac{2}{3} [\lim_{p \to \infty} \frac{x^{2} + 1}{x + \sqrt{x^{2} + 1}} - 1] = - \frac{2}{3} [0 - 1]$ $= \frac{2}{3}$
	Answered by Dwaipayan Shikari last updated on 04/Dec/20

	$\frac{1}{x + \sqrt{x^{2} + 1}} = t \Rightarrow - \frac{1}{(\sqrt{x^{2} + 1}) (x + \sqrt{x^{2} + 1})} = \frac{d t}{d x}$ $= - \int \sqrt{x^{2} + 1} t d t$ $= - \int t \sqrt{{(\frac{t^{2} - 1}{2 t})}^{2} + 1} d t$ $= - \frac{1}{2} \int \sqrt{{(t^{2} + 1)}^{2}} = - \frac{1}{6} t^{3} - \frac{t}{2} = - \frac{1}{6} {(\frac{1}{x + \sqrt{x^{2} + 1}})}^{3} - \frac{1}{2} (\frac{1}{x + \sqrt{x^{2} + 1}})$ $= \frac{1}{6} {(x - \sqrt{x^{2} + 1})}^{3} + \frac{1}{2} (x - \sqrt{x^{2} + 1}) + C$
	Answered by Ar Brandon last updated on 04/Dec/20
	$I=∫(dx/((x+(√(x^2 +1)))^2 )) t=x+(√(x^2 +1)) , x=((t^2 −1)/(2t)), dx=((t^2 +1)/(2t^2 ))dt I=∫(1/t^2 )∙((t^2 +1)/(2t^2 ))dt=(1/2)∫{(1/t^2 )+(1/t^4 )}dt =−(1/2){(1/t)+(1/(3t^3 ))}=−(1/2){(1/(x+(√(x^2 +1))))+(1/3)∙(1/((x+(√(x^2 +1)))^3 ))+C}$
	$I = \int \frac{dx}{{(x + \sqrt{x^{2} + 1})}^{2}}$ $t = x + \sqrt{x^{2} + 1}, x = \frac{t^{2} - 1}{2 t}, dx = \frac{t^{2} + 1}{{2 t}^{2}} dt$ $I = \int \frac{1}{t^{2}} \cdot \frac{t^{2} + 1}{{2 t}^{2}} dt = \frac{1}{2} \int {\frac{1}{t^{2}} + \frac{1}{t^{4}}} dt$ $= - \frac{1}{2} {\frac{1}{t} + \frac{1}{{3 t}^{3}}} = - \frac{1}{2} {\frac{1}{x + \sqrt{x^{2} + 1}} + \frac{1}{3} \cdot \frac{1}{{(x + \sqrt{x^{2} + 1})}^{3}} + C}$
	Answered by Ar Brandon last updated on 04/Dec/20
	$I=∫(dx/((x+(√(x^2 +1)))^2 )) x=tanθ ⇒ dx=sec^2 θdθ I=∫((sec^2 θdθ)/((tanθ+(√(tan^2 θ+1)))^2 ))=∫((sec^2 θdθ)/((tanθ+secθ)^2 )) =∫(dθ/((sinθ+1)^2 ))=∫(((1−sinθ)^2 )/((1−sin^2 θ)^2 ))dθ =∫{((1−2sinθ+sin^2 θ)/(cos^4 θ))}dθ=∫{((cos^2 θ−2sinθ+2sin^2 θ)/(cos^4 θ))}dθ =∫sec^2 θdθ+2∫(((−sinθ))/(cos^4 θ))dθ+2∫tan^2 θsec^2 θdθ =tanθ−(2/(3cos^3 θ))+((2tan^3 θ)/3)+C where θ=tan^(−1) (x)$
	$I = \int \frac{dx}{{(x + \sqrt{x^{2} + 1})}^{2}}$ $x = \tan θ \Rightarrow dx = \sec^{2} θ d θ$ $I = \int \frac{\sec^{2} θ d θ}{{(\tan θ + \sqrt{\tan^{2} θ + 1})}^{2}} = \int \frac{\sec^{2} θ d θ}{{(\tan θ + \sec θ)}^{2}}$ $= \int \frac{d θ}{{(\sin θ + 1)}^{2}} = \int \frac{{(1 - \sin θ)}^{2}}{{(1 - \sin^{2} θ)}^{2}} d θ$ $= \int {\frac{1 - 2 \sin θ + \sin^{2} θ}{\cos^{4} θ}} d θ = \int {\frac{\cos^{2} θ - 2 \sin θ + {2 \sin}^{2} θ}{\cos^{4} θ}} d θ$ $= \int \sec^{2} θ d θ + 2 \int \frac{(- \sin θ)}{\cos^{4} θ} d θ + 2 \int \tan^{2} θ \sec^{2} θ d θ$ $= \tan θ - \frac{2}{{3 \cos}^{3} θ} + \frac{{2 \tan}^{3} θ}{3} + C$ $w h e r e θ = \tan^{- 1} (x)$
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