find lim_(x→0^+ ) ∫_x ^(2x) ((√(1+t^2 ))/t)dt .


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Question Number 44319 by abdo.msup.com last updated on 26/Sep/18

$f i n d {l i m}_{x \to 0^{+}} \int_{x}^{2 x} \frac{\sqrt{1 + t^{2}}}{t} d t .$
Commented by maxmathsup by imad last updated on 27/Sep/18
$changement t =sh(u) give ∫_(argsh(x)) ^(argsh(2x)) ((√(1+sh^2 u))/(sh(u))) ch(u)dt =∫_(argsh(x)) ^(argsh(2x)) ((ch^2 u)/(sh(u)))dt = ∫_(ln(x+(√(1+x^2 )))) ^(ln(2x+(√(1+4x^2 )))) ((1+ch(2u))/(e^u −e^(−u) ))du =∫_(ln(x+(√(1+x^2 )))) ^(ln(2x+(√(1+4x^2 )))) ((1+((e^(2u) +e^(−2u) )/2))/(e^u −e^(−u) ))du =(1/2)∫_(ln(x+(√(1+x^2 )))) ^(ln(2x +(√(1+4x^2 ))) ((2 +e^(2u) +e^(−2u) )/(e^u −e^(−u) )) du =_(e^u =α) (1/2) ∫_(x+(√(1+x^2 ))) ^(2x +(√(1+4x^2 ))) ((2 +α^2 +α^(−2) )/(α−α^(−1) ))dα =(1/2) ∫_(x+(√(1+x^2 ))) ^(2x +(√(1+4x^2 ))) ((2α^2 +α^4 +1)/(α^3 −α))dα =(1/2) ∫_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) ((α^4 +2α^2 +1)/(α^3 −α)) dα=(1/2) ∫_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) ((α(α^3 −α)+3α^2 +1)/(α^3 −α))dα =(1/2)[(α^2 /2)]_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) +(1/2) ∫_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) ((3α^2 +1)/(α^3 −α))dα =(1/4){ (2x+(√(1+4x^2 )))^2 −(x+(√(1+x^2 )))^2 } +(1/2) ∫_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) ((3α^2 +1)/(α^3 −α)) dα let decompose F(α) =((3α^2 +1)/(α^3 −α)) =((3α^2 +1)/(α(α−1)(α+1))) F(α) = (a/α) +(b/(α−1)) +(c/(α+1)) ⇒a =−1 b =(4/2) =2 and c=(4/2) =2 ⇒F(α) =−(1/α) +(2/(α−1)) +(2/(α+1)) ⇒ ∫_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) F(α)dα = [2ln∣α−1∣ +2 ln∣α+1∣−ln∣α∣]_(x+(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) =2ln∣2x−1+(√(1+4x^2 )) ∣+2ln∣2x+1+(√(1+4x^2 ∣))−ln∣2x+(√(1+4x^2 ))∣ −2ln∣x−1 +(√(1+x^2 ))∣ −2ln∣x+1 +(√(1+x^2 ))∣+ln∣x+(√(1+x^2 )) =2ln∣ ((2x−1+(√(1+4x^2 )))/(x−1 +(√(1+x^2 ))))∣ +2ln∣((2x+1+(√(1+4x^2 )))/(x+1 +(√(1+x^2 ))))∣ −ln∣((2x+(√(1+4x^2 )))/(x+(√(1+x^2 )))) ∣ but lim_(x→0) ((2x−1 +(√(1+4x^2 )))/(x−1+(√(1+x^2 )))) =lim_(x→0) ((2 +((8x)/(2(√(1+4x^2 )))))/(1+ ((2x)/(2(√(1+x^2 )))))) =2 ⇒ (1/2) ∫_(x +(√(1+x^2 ))) ^(2x+(√(1+4x^2 ))) F(α) →ln(2) (x→0) ⇒ lim_(x→0^+ ) ∫_x ^(2x) ((√(1+x^2 ))/x) dx =ln(2) .$
$c h a n g e m e n t t = s h (u) g i v e \int_{a r g s h (x)}^{a r g s h (2 x)} \frac{\sqrt{1 + {s h}^{2} u}}{s h (u)} c h (u) d t$ $= \int_{a r g s h (x)}^{a r g s h (2 x)} \frac{{c h}^{2} u}{s h (u)} d t = \int_{l n (x + \sqrt{1 + x^{2})}}^{l n (2 x + \sqrt{1 + 4 x^{2})}} \frac{1 + c h (2 u)}{e^{u} - e^{- u}} d u$ $= \int_{l n (x + \sqrt{1 + x^{2})}}^{l n (2 x + \sqrt{1 + 4 x^{2}})} \frac{1 + \frac{e^{2 u} + e^{- 2 u}}{2}}{e^{u} - e^{- u}} d u = \frac{1}{2} \int_{l n (x + \sqrt{1 + x^{2})}}^{l n (2 x + \sqrt{1 + 4 x^{2}}} \frac{2 + e^{2 u} + e^{- 2 u}}{e^{u} - e^{- u}} d u$ $=_{e^{u} = α} \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{2 + α^{2} + α^{- 2}}{α - α^{- 1}} d α = \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{2 α^{2} + α^{4} + 1}{α^{3} - α} d α$ $= \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{α^{4} + 2 α^{2} + 1}{α^{3} - α} d α = \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{α (α^{3} - α) + 3 α^{2} + 1}{α^{3} - α} d α$ $= \frac{1}{2} {[\frac{α^{2}}{2}]}_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} + \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{3 α^{2} + 1}{α^{3} - α} d α$ $= \frac{1}{4} {{(2 x + \sqrt{1 + 4 x^{2}})}^{2} - {(x + \sqrt{1 + x^{2}})}^{2}} + \frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} \frac{3 α^{2} + 1}{α^{3} - α} d α$ $l e t d e c o m p o s e F (α) = \frac{3 α^{2} + 1}{α^{3} - α} = \frac{3 α^{2} + 1}{α (α - 1) (α + 1)}$ $F (α) = \frac{a}{α} + \frac{b}{α - 1} + \frac{c}{α + 1} \Rightarrow a = - 1$ $b = \frac{4}{2} = 2 a n d c = \frac{4}{2} = 2 \Rightarrow F (α) = - \frac{1}{α} + \frac{2}{α - 1} + \frac{2}{α + 1} \Rightarrow$ $\int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} F (α) d α = {[2 l n ∣ α - 1 ∣ + 2 l n ∣ α + 1 ∣ - l n ∣ α ∣]}_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}}$ $= 2 l n ∣ 2 x - 1 + \sqrt{1 + 4 x^{2}} ∣ + 2 l n ∣ 2 x + 1 + \sqrt{1 + 4 x^{2} ∣} - l n ∣ 2 x + \sqrt{1 + 4 x^{2}} ∣$ $- 2 l n ∣ x - 1 + \sqrt{1 + x^{2}} ∣ - 2 l n ∣ x + 1 + \sqrt{1 + x^{2}} ∣ + l n ∣ x + \sqrt{1 + x^{2}}$ $= 2 l n ∣ \frac{2 x - 1 + \sqrt{1 + 4 x^{2}}}{x - 1 + \sqrt{1 + x^{2}}} ∣ + 2 l n ∣ \frac{2 x + 1 + \sqrt{1 + 4 x^{2}}}{x + 1 + \sqrt{1 + x^{2}}} ∣ - l n ∣ \frac{2 x + \sqrt{1 + 4 x^{2}}}{x + \sqrt{1 + x^{2}}} ∣$ $b u t {l i m}_{x \to 0} \frac{2 x - 1 + \sqrt{1 + 4 x^{2}}}{x - 1 + \sqrt{1 + x^{2}}} = {l i m}_{x \to 0} \frac{2 + \frac{8 x}{2 \sqrt{1 + 4 x^{2}}}}{1 + \frac{2 x}{2 \sqrt{1 + x^{2}}}} = 2 \Rightarrow$ $\frac{1}{2} \int_{x + \sqrt{1 + x^{2}}}^{2 x + \sqrt{1 + 4 x^{2}}} F (α) \to l n (2) (x \to 0) \Rightarrow$ ${l i m}_{x \to 0^{+}} \int_{x}^{2 x} \frac{\sqrt{1 + x^{2}}}{x} d x = l n (2) .$
Commented by maxmathsup by imad last updated on 27/Sep/18

$a n o t h e r b u t e a s y \exists c \in] x, 2 x [/ \int_{x}^{2 x} \frac{\sqrt{1 + t^{2}}}{t} d t = \sqrt{1 + c^{2}} \int_{x}^{2 x} \frac{d t}{t}$ $= \sqrt{1 + c^{2}} l n ∣ \frac{2 x}{x} ∣ b u t x \to 0 \Rightarrow c \to 0 {l i m}_{x \to 0} \int_{x}^{2 x} \frac{\sqrt{1 + t^{2}}}{t} d t = l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18
	$y^2 =1+t^2 2ydy=2tdt ∫((t(√(1+t^2 )))/t^2 )dt ∫((y×ydy)/(y^2 −1)) ∫((y^2 −1+1)/(y^2 −1))dy ∫dy+∫(dy/((y^ +1)(y−1))) ∫dy+(1/2)∫(((y+1)−(y−1))/((y+1)(y−1)))dy y+(1/2)ln(((y−1)/(y+1))) (√(t^2 +1)) +(1/2)ln((((√(t^2 +1)) −1)/((√(t^2 +1)) +1))) ∫_x ^(2x) ((√(1+t^2 ))/t)dt {(√(4x^2 +1)) −(√(x^2 +1)) }+(1/2){ln((((√(4x^2 +1)) −1)/((√(4x^2 +1)) +1)))−ln((((√(x^2 +1)) −1)/((√(x^2 +1)) +1)))} when lim x→0+ first expression=0 but second expression (1/2){ln((0/2))−ln((0/2))} pls check... i$
	$y^{2} = 1 + t^{2} 2 y d y = 2 t d t$ $\int \frac{t \sqrt{1 + t^{2}}}{t^{2}} d t$ $\int \frac{y \times y d y}{y^{2} - 1}$ $\int \frac{y^{2} - 1 + 1}{y^{2} - 1} d y$ $\int d y + \int \frac{d y}{(y^{} + 1) (y - 1)}$ $\int d y + \frac{1}{2} \int \frac{(y + 1) - (y - 1)}{(y + 1) (y - 1)} d y$ $y + \frac{1}{2} l n (\frac{y - 1}{y + 1})$ $\sqrt{t^{2} + 1} + \frac{1}{2} l n (\frac{\sqrt{t^{2} + 1} - 1}{\sqrt{t^{2} + 1} + 1})$ $\int_{x}^{2 x} \frac{\sqrt{1 + t^{2}}}{t} d t$ ${\sqrt{4 x^{2} + 1} - \sqrt{x^{2} + 1}} + \frac{1}{2} {l n (\frac{\sqrt{4 x^{2} + 1} - 1}{\sqrt{4 x^{2} + 1} + 1}) - l n (\frac{\sqrt{x^{2} + 1} - 1}{\sqrt{x^{2} + 1} + 1})}$ $w h e n l i m x \to 0 +$ $f i r s t e x p r e s s i o n = 0$ $b u t s e c o n d e x p r e s s i o n \frac{1}{2} {l n (\frac{0}{2}) - l n (\frac{0}{2})}$ $p l s c h e c k . . .$ $i$
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