calculate ∫_(√3) ^(+∞) (dx/(x(√( 2+x^2 )))) .


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Question Number 35686 by prof Abdo imad last updated on 22/May/18

$c a l c u l a t e \int_{\sqrt{3}}^{+ \infty} \frac{d x}{x \sqrt{2 + x^{2}}} .$
Commented by prof Abdo imad last updated on 22/May/18
$let put I = ∫_(√3) ^(+∞) (dx/(x(√(2+x^2 )))) changement x=(√2)sh(t) give t =argsh((x/(√2))) and I = ∫_(argsh(((√3)/(√2)))) ^(+∞) (((√2)ch(t)dt)/((√2)sh(t)(√2)(√(1+sht^2 )))) = ∫_(ln(((√3)/(√2)) +(√(1 +(3/2))))) ^(+∞) (dt/((√2)sh(t))) =(1/(√2)) ∫_(ln( (((√3) +(√5))/(√2)))) ^(+∞) 2(dt/(e^t −e^(−t) )) =(√2) ∫_(ln((((√3) +(√5))/(√2)))) ^(+∞) (dt/(e^t −e^(−t) )) =_(e^t =x) (√2) ∫_(((√3)+(√5))/(√2)) ^(+∞) (1/(x−(1/x))) (dx/x) I = (√2) ∫_(((√3) +(√5))/(√2)) ^(+∞) (dx/(x^2 −1)) =((√2)/2) ∫_(((√3)+(√5))/(√2)) ^(+∞) { (1/(x−1)) −(1/(x+1))}dx =((√2)/2)[ln∣((x−1)/(x+1))∣]_(((√3) +(√5))/(√2)) ^(+∞) =−((√2)/2)ln((((((√3)+(√5))/(√2))−1)/((((√3)+(√5))/(√2))+1))) =−((√2)/2) ln((((√3) +(√5) −(√2))/((√3) +(√5) +(√2))))$
$l e t p u t I = \int_{\sqrt{3}}^{+ \infty} \frac{d x}{x \sqrt{2 + x^{2}}} c h a n g e m e n t$ $x = \sqrt{2} s h (t) g i v e t = a r g s h (\frac{x}{\sqrt{2}}) a n d$ $I = \int_{a r g s h (\frac{\sqrt{3}}{\sqrt{2}})}^{+ \infty} \frac{\sqrt{2} c h (t) d t}{\sqrt{2} s h (t) \sqrt{2} \sqrt{1 + {s h t}^{2}}}$ $= \int_{l n (\frac{\sqrt{3}}{\sqrt{2}} + \sqrt{1 + \frac{3}{2}})}^{+ \infty} \frac{d t}{\sqrt{2} s h (t)}$ $= \frac{1}{\sqrt{2}} \int_{l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{+ \infty} 2 \frac{d t}{e^{t} - e^{- t}}$ $= \sqrt{2} \int_{l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{+ \infty} \frac{d t}{e^{t} - e^{- t}}$ $=_{e^{t} = x} \sqrt{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} \frac{1}{x - \frac{1}{x}} \frac{d x}{x}$ $I = \sqrt{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} \frac{d x}{x^{2} - 1}$ $= \frac{\sqrt{2}}{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} {\frac{1}{x - 1} - \frac{1}{x + 1}} d x$ $= \frac{\sqrt{2}}{2} {[l n ∣ \frac{x - 1}{x + 1} ∣]}_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} = - \frac{\sqrt{2}}{2} l n (\frac{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}} - 1}{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}} + 1})$ $= - \frac{\sqrt{2}}{2} l n (\frac{\sqrt{3} + \sqrt{5} - \sqrt{2}}{\sqrt{3} + \sqrt{5} + \sqrt{2}})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18
	$=∫_(√3) ^∞ ((xdx)/(x^2 (√(2+x^2 )))) t^2 =2+x^2 2tdt=2xdx =∫_(√5) ^∞ ((tdt)/((t^2 −2)t)) =∫_(√5) ^∞ (dt/((t^2 −2))) =(1/(2(√2)))∣ln(((t−(√2))/(t+(√2))))∣_(√5) ^∞ =(1/(2(√2)))∣ln(((1−(((√2) )/t))/(1+(((√2) )/t))))∣_(√5) ^∞ =(1/(2(√2))){0−ln(((1−((√2)/((√5) )))/(1+((√(2 ))/(√5)))))} =(1/(2(√2))){−ln((((√5) −(√2))/(√(5+(√2) _ ))) }$
	$= \int_{\sqrt{3}}^{\infty} \frac{x d x}{x^{2} \sqrt{2 + x^{2}}}$ $t^{2} = 2 + x^{2} 2 t d t = 2 x d x$ $= \int_{\sqrt{5}}^{\infty} \frac{t d t}{(t^{2} - 2) t}$ $= \int_{\sqrt{5}}^{\infty} \frac{d t}{(t^{2} - 2)}$ $= \frac{1}{2 \sqrt{2}} ∣ l n (\frac{t - \sqrt{2}}{t + \sqrt{2}}) ∣_{\sqrt{5}}^{\infty}$ $= \frac{1}{2 \sqrt{2}} ∣ l n (\frac{1 - \frac{\sqrt{2}}{t}}{1 + \frac{\sqrt{2}}{t}}) ∣_{\sqrt{5}}^{\infty}$ $= \frac{1}{2 \sqrt{2}} {0 - l n (\frac{1 - \frac{\sqrt{2}}{\sqrt{5}}}{1 + \frac{\sqrt{2}}{\sqrt{5}}})}$ $= \frac{1}{2 \sqrt{2}} {- l n (\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5 + \sqrt{2}_{}}}}$
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