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Question Number 40467 by gunawan last updated on 22/Jul/18

∫ln ∣(√(x+1))+(√x)∣ dx=

$$\int\mathrm{ln}\:\mid\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\mid\:{dx}= \\ $$

Commented by maxmathsup by imad last updated on 22/Jul/18

let I = ∫ ln∣(√(x+1)) +(√x)∣dx  due to x≥0 we have  I = ∫ ln((√(x+1))+(√x))dx =_((√x)=t)  2  ∫ln((√(1+t^2 )) +t) t dt  by parts  u =ln(t+(√(1+t^2 )))  and   v^′  =t ⇒  I  = 2{  (t^2 /2)ln(t+(√(1+t^2 ))) −∫  (t^2 /2) (1/(√(1+t^2 )))dt}  =t^2 ln(t+(√(1+t^2 ))) −∫   ((t^2 +1−1)/(√(1+t^2 )))dt  =t^2 ln(t+(√(1+t^2 ))) −∫  (√(1+t^2 ))dt   + ∫   (dt/(√(1+t^2 )))  =t^2 ln(t+(√(1+t^2 ))) +ln(t+(√(1+t^2 )))  − ∫(√(1+t^2 ))dt  ∫  (√(1+t^2 ))dt =_(t=sh(x))    ∫ch(x)ch(x)= ∫ ((1+ch(2x))/2)dx  =(1/2) x  +(1/4)sh(2x)=(x/2) +(1/2)sh(x)ch(x)  =(1/2)ln(t+(√(1+t^2 )))   +(1/2)t(√(1+t^2 ))  ⇒  I  =  (t^2  +(3/2))ln(t +(√(1+t^2 )))  +((t(√(1+t^2 )))/2)  +c  .

$${let}\:{I}\:=\:\int\:{ln}\mid\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\mid{dx}\:\:{due}\:{to}\:{x}\geqslant\mathrm{0}\:{we}\:{have} \\ $$$${I}\:=\:\int\:{ln}\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\right){dx}\:=_{\sqrt{{x}}={t}} \:\mathrm{2}\:\:\int{ln}\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:+{t}\right)\:{t}\:{dt}\:\:{by}\:{parts} \\ $$$${u}\:={ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:\:{and}\:\:\:{v}^{'} \:={t}\:\Rightarrow \\ $$$${I}\:\:=\:\mathrm{2}\left\{\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:−\int\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}\right\} \\ $$$$={t}^{\mathrm{2}} {ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:−\int\:\:\:\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$={t}^{\mathrm{2}} {ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:−\int\:\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:+\:\int\:\:\:\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$={t}^{\mathrm{2}} {ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:+{ln}\left({t}+\sqrt{\left.\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\:−\:\int\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right. \\ $$$$\int\:\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=_{{t}={sh}\left({x}\right)} \:\:\:\int{ch}\left({x}\right){ch}\left({x}\right)=\:\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{x}\:\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{x}\right)=\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({x}\right){ch}\left({x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}+\sqrt{\left.\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\Rightarrow\right. \\ $$$${I}\:\:=\:\:\left({t}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{2}}\right){ln}\left({t}\:+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:\:+\frac{{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}}\:\:+{c}\:\:. \\ $$

Commented by math khazana by abdo last updated on 22/Jul/18

⇒ I =(x+(3/2))ln((√x)+(√(1+x)))  +(((√x)(√(1+x)))/2) +c .

$$\Rightarrow\:{I}\:=\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right){ln}\left(\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}\right)\:\:+\frac{\sqrt{{x}}\sqrt{\mathrm{1}+{x}}}{\mathrm{2}}\:+{c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jul/18

use ∫uvdx  ln((√(x+1)) +(√x) )x−∫(((1/(2(√(x+1))))+(1/(2(√x) )))/((√(x+1)) +(√x) )).xdx  xln((√(x+1)) +(√x) )−(1/2)∫(((√(x+1)) +(√x) )/(((√(x+1)) +(√(x))) (√(x+1)) .(√x)))xdx  xln((√(x+1)) +(√x) )−(1/2)∫(((√x) )/(√(x+1)))dx  let t^2 =x+1  dx=2tdt  I_2 =∫(((√x) )/((√(x+1)) ))dx   =∫(((√(t^2 −1)) )/t)2tdt  2∫(√(t^2 −1)) dt  2{(t/2)(√(t^2 −1)) −(1/2)ln(t+(√(t^2 −1)) }  {(√(x^2 −1)) .x−ln((√(x^2 −1)) +x)}  now rearrannge by yourself...

$${use}\:\int{uvdx} \\ $$$${ln}\left(\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:\right){x}−\int\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\:}}{\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:}.{xdx} \\ $$$${xln}\left(\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:}{\left(\sqrt{{x}+\mathrm{1}}\:+\sqrt{\left.{x}\right)}\:\sqrt{{x}+\mathrm{1}}\:.\sqrt{{x}}\right.}{xdx} \\ $$$${xln}\left(\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{{x}}\:}{\sqrt{{x}+\mathrm{1}}}{dx} \\ $$$${let}\:{t}^{\mathrm{2}} ={x}+\mathrm{1}\:\:{dx}=\mathrm{2}{tdt} \\ $$$${I}_{\mathrm{2}} =\int\frac{\sqrt{{x}}\:}{\sqrt{{x}+\mathrm{1}}\:}{dx} \\ $$$$\:=\int\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:}{{t}}\mathrm{2}{tdt} \\ $$$$\mathrm{2}\int\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:{dt} \\ $$$$\mathrm{2}\left\{\frac{{t}}{\mathrm{2}}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\right\}\right. \\ $$$$\left\{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:.{x}−{ln}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+{x}\right)\right\} \\ $$$${now}\:{rearrannge}\:{by}\:{yourself}... \\ $$

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