Question Number 44319 by abdo.msup.com last updated on 26/Sep/18
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}{dt}\:. \\ $$
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) .](Q44344.png)
$${changement}\:{t}\:={sh}\left({u}\right)\:{give}\:\:\int_{{argsh}\left({x}\right)} ^{{argsh}\left(\mathrm{2}{x}\right)} \:\:\frac{\sqrt{\mathrm{1}+{sh}^{\mathrm{2}} {u}}}{{sh}\left({u}\right)}\:{ch}\left({u}\right){dt} \\ $$$$=\int_{{argsh}\left({x}\right)} ^{{argsh}\left(\mathrm{2}{x}\right)} \:\frac{{ch}^{\mathrm{2}} {u}}{{sh}\left({u}\right)}{dt}\:\:=\:\:\int_{{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.} ^{{ln}\left(\mathrm{2}{x}+\sqrt{\left.\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}\right.} \:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{{e}^{{u}} \:−{e}^{−{u}} }{du} \\ $$$$=\int_{{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.} ^{{ln}\left(\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)} \:\:\:\frac{\mathrm{1}+\frac{{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} }{\mathrm{2}}}{{e}^{{u}} \:−{e}^{−{u}} }{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.} ^{{ln}\left(\mathrm{2}{x}\:+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right.} \:\:\frac{\mathrm{2}\:+{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} }{{e}^{{u}} \:−{e}^{−{u}} }\:{du} \\ $$$$=_{{e}^{{u}} =\alpha} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}\:+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\:\frac{\mathrm{2}\:+\alpha^{\mathrm{2}} \:+\alpha^{−\mathrm{2}} }{\alpha−\alpha^{−\mathrm{1}} }{d}\alpha\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}\:+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\frac{\mathrm{2}\alpha^{\mathrm{2}} \:+\alpha^{\mathrm{4}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} \:−\alpha}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\:\frac{\alpha^{\mathrm{4}} \:+\mathrm{2}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} \:−\alpha}\:{d}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \frac{\alpha\left(\alpha^{\mathrm{3}} −\alpha\right)+\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} −\alpha}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}\right]_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\:\frac{\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} −\alpha}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\left(\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:−\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \right\}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\:\:\frac{\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} \:−\alpha}\:{d}\alpha \\ $$$${let}\:{decompose}\:{F}\left(\alpha\right)\:=\frac{\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha^{\mathrm{3}} \:−\alpha}\:=\frac{\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\alpha\left(\alpha−\mathrm{1}\right)\left(\alpha+\mathrm{1}\right)} \\ $$$${F}\left(\alpha\right)\:=\:\frac{{a}}{\alpha}\:+\frac{{b}}{\alpha−\mathrm{1}}\:+\frac{{c}}{\alpha+\mathrm{1}}\:\Rightarrow{a}\:=−\mathrm{1}\:\:\: \\ $$$${b}\:=\frac{\mathrm{4}}{\mathrm{2}}\:=\mathrm{2}\:{and}\:\:\:\:\:{c}=\frac{\mathrm{4}}{\mathrm{2}}\:=\mathrm{2}\:\Rightarrow{F}\left(\alpha\right)\:=−\frac{\mathrm{1}}{\alpha}\:+\frac{\mathrm{2}}{\alpha−\mathrm{1}}\:+\frac{\mathrm{2}}{\alpha+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:{F}\left(\alpha\right){d}\alpha\:=\:\left[\mathrm{2}{ln}\mid\alpha−\mathrm{1}\mid\:+\mathrm{2}\:{ln}\mid\alpha+\mathrm{1}\mid−{ln}\mid\alpha\mid\right]_{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \\ $$$$=\mathrm{2}{ln}\mid\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\mid+\mathrm{2}{ln}\mid\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \mid}−{ln}\mid\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\mid \\ $$$$−\mathrm{2}{ln}\mid{x}−\mathrm{1}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\:−\mathrm{2}{ln}\mid{x}+\mathrm{1}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid+{ln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\: \\ $$$$=\mathrm{2}{ln}\mid\:\frac{\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{{x}−\mathrm{1}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid\:+\mathrm{2}{ln}\mid\frac{\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{{x}+\mathrm{1}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid\:−{ln}\mid\frac{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\mid \\ $$$${but}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\mathrm{2}{x}−\mathrm{1}\:+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{{x}−\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}\:+\frac{\mathrm{8}{x}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}}{\mathrm{1}+\:\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}\:=\mathrm{2}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ^{\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }} \:\:\:\:{F}\left(\alpha\right)\:\rightarrow{ln}\left(\mathrm{2}\right)\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:{dx}\:={ln}\left(\mathrm{2}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 27/Sep/18
![another but easy ∃ c ∈]x,2x[ / ∫_x ^(2x) ((√(1+t^2 ))/t)dt =(√(1+c^2 )) ∫_x ^(2x) (dt/t) =(√(1+c^2 ))ln∣((2x)/x)∣ but x→0 ⇒ c→0 lim_(x→0) ∫_x ^(2x) ((√(1+t^2 ))/t)dt =ln(2).](Q44345.png)
$$\left.{another}\:{but}\:{easy}\:\:\:\:\exists\:{c}\:\in\right]{x},\mathrm{2}{x}\left[\:\:/\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}{dt}\:=\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }\:\int_{{x}} ^{\mathrm{2}{x}} \:\frac{{dt}}{{t}}\right. \\ $$$$=\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }{ln}\mid\frac{\mathrm{2}{x}}{{x}}\mid\:\:\:{but}\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{c}\rightarrow\mathrm{0}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\int_{{x}} ^{\mathrm{2}{x}} \:\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}{dt}\:={ln}\left(\mathrm{2}\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18
$${y}^{\mathrm{2}} =\mathrm{1}+{t}^{\mathrm{2}} \:\:\mathrm{2}{ydy}=\mathrm{2}{tdt} \\ $$$$\int\frac{{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{{y}×{ydy}}{{y}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\int\frac{{y}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{y}^{\mathrm{2}} −\mathrm{1}}{dy} \\ $$$$\int{dy}+\int\frac{{dy}}{\left({y}^{} +\mathrm{1}\right)\left({y}−\mathrm{1}\right)} \\ $$$$\int{dy}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({y}+\mathrm{1}\right)−\left({y}−\mathrm{1}\right)}{\left({y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)}{dy} \\ $$$${y}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}}\right) \\ $$$$\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{1}}\right) \\ $$$$\int_{{x}} ^{\mathrm{2}{x}} \frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}{dt} \\ $$$$\left\{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{1}}\right)−{ln}\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{1}}\right)\right\} \\ $$$$\:\:{when}\:{lim}\:{x}\rightarrow\mathrm{0}+\: \\ $$$${first}\:{expression}=\mathrm{0} \\ $$$${but}\:{second}\:{expression}\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{0}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{0}}{\mathrm{2}}\right)\right\} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$${i} \\ $$