find ∫ (dx/((x+1)(√(1+x^2 ))))


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Question Number 37067 by math khazana by abdo last updated on 08/Jun/18

$f i n d \int \frac{d x}{(x + 1) \sqrt{1 + x^{2}}}$
Commented by math khazana by abdo last updated on 09/Jun/18
$changement x+1=(1/t) give t=(1/(x+1)) I = ∫ (t/(√(1+((1/t)−1)^2 ))) ((−dt)/t^2 ) =−∫ (dt/(t(√(1+ (((t−1)^2 )/t^2 ))))) =−∫ ((t dt)/(t(√(t^2 +t^2 −2t +1)))) =−∫ (dt/(√(2t^2 −2t +1)))?but 2t^2 −2t +1 =2{ t^2 −t +(1/2)} =2 { (t−(1/2))^2 +(1/2) −(1/4)}=2{ (t−(1/2))^2 +(1/4)} changemnt t−(1/2)=(1/2) u?give I = −(1/(√2)) ∫ (1/((1/2)(√(1+u^2 )))) (du/2) =−(1/(√2)) ∫ (du/(√(1+u^2 ))) +c =−(1/(√2)) ln(u +(√(1+u^2 )) )+c = −(1/(√2))ln( 2t−1 +(√(1+(2t−1)^2 ))) +c =−(1/(√2))ln{ (2/(x+1)) −1 +(√(1+((2/(x+1)) −1)^2 )) +c$
$c h a n g e m e n t x + 1 = \frac{1}{t} g i v e t = \frac{1}{x + 1}$ $I = \int \frac{t}{\sqrt{1 + {(\frac{1}{t} - 1)}^{2}}} \frac{- d t}{t^{2}}$ $= - \int \frac{d t}{t \sqrt{1 + \frac{{(t - 1)}^{2}}{t^{2}}}}$ $= - \int \frac{t d t}{t \sqrt{t^{2} + t^{2} - 2 t + 1}}$ $= - \int \frac{d t}{\sqrt{2 t^{2} - 2 t + 1}} ? b u t$ $2 t^{2} - 2 t + 1 = 2 {t^{2} - t + \frac{1}{2}}$ $= 2 {{(t - \frac{1}{2})}^{2} + \frac{1}{2} - \frac{1}{4}} = 2 {{(t - \frac{1}{2})}^{2} + \frac{1}{4}}$ $c h a n g e m n t t - \frac{1}{2} = \frac{1}{2} u ? g i v e$ $I = - \frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{2} \sqrt{1 + u^{2}}} \frac{d u}{2}$ $= - \frac{1}{\sqrt{2}} \int \frac{d u}{\sqrt{1 + u^{2}}} + c$ $= - \frac{1}{\sqrt{2}} l n (u + \sqrt{1 + u^{2}}) + c$ $= - \frac{1}{\sqrt{2}} l n (2 t - 1 + \sqrt{1 + {(2 t - 1)}^{2}}) + c$ $= - \frac{1}{\sqrt{2}} l n {\frac{2}{x + 1} - 1 + \sqrt{1 + {(\frac{2}{x + 1} - 1)}^{2}} + c$
	Answered by MJS last updated on 08/Jun/18

	$see my answer to qu . 37018 (1)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

	$t = \frac{1}{x + 1}$ $x + 1 = \frac{1}{t} d x = - \frac{1}{t^{2}} d t$ $= - \int \frac{d t}{t^{2}} \times \frac{t}{\sqrt{1 + {(\frac{1}{t} - 1)}^{2}}}$ $= - \int \frac{d t}{t} \times \frac{t}{\sqrt{t^{2} + (1 - 2 t + t^{2})}}$ $= - \int \frac{d t}{\sqrt{2 t^{2} - 2 t + 1}}$ $= - \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{t^{2} - t + \frac{1}{2}}}$ $= \frac{- 1}{\sqrt{2}} \int \frac{d t}{\sqrt{t^{2} - 2 . t . \frac{1}{2} + \frac{1}{4} + \frac{1}{2} - \frac{1}{4}}}$ $= \frac{- 1}{\sqrt{2}} \int \frac{d t}{\sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2}}}$ $n o w u s e f o r m u l a \int \frac{d x}{\sqrt{x^{2} + a^{2}}}$
	Commented by math khazana by abdo last updated on 09/Jun/18

	$s i r T a n m a y y o u h a v e c o m m i t e d a e r r o r o f$ $c a l c u l u s b e c a u s e t^{2} - t + \frac{1}{2} = {(t - \frac{1}{2})}^{2} + \frac{1}{4}!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

	$y e s i n p l a c e o f \frac{1}{2} i p u t 1 . . . r e c t i f y i n g i n r e d c o l o u r$
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