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Question Number 62129 by maxmathsup by imad last updated on 15/Jun/19
let f(x)=ln(x+1−2(√x))  1) find D_f   2) determine f^(−1)   3) calculate  ∫f (x)dx and  ∫ f^(−1) (x)dx
$${let}\:{f}\left({x}\right)={ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{−\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int{f}\:\left({x}\right){dx}\:{and}\:\:\int\:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$
Commented by arcana last updated on 16/Jun/19
1)  x+1−2(√x)>0⇒((√x)−1)^2 >0  necessary (√x)−1>0 ∨ (√x)−1<0  (√x)>1∨(√x)<1  x>1∨x<1, x≥0  ⇒D_f =[0,1)∪(1,+∞)=R^+ −{1}
$$\left.\mathrm{1}\right) \\ $$$${x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}>\mathrm{0}\Rightarrow\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$${necessary}\:\sqrt{{x}}−\mathrm{1}>\mathrm{0}\:\vee\:\sqrt{{x}}−\mathrm{1}<\mathrm{0} \\ $$$$\sqrt{{x}}>\mathrm{1}\vee\sqrt{{x}}<\mathrm{1} \\ $$$${x}>\mathrm{1}\vee{x}<\mathrm{1},\:{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow{D}_{{f}} =\left[\mathrm{0},\mathrm{1}\right)\cup\left(\mathrm{1},+\infty\right)=\mathbb{R}^{+} −\left\{\mathrm{1}\right\} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 16/Jun/19
e^y =((√x)−1)^2 ⇒(√e^y )=(√x)−1⇒1+(√e^y )=(√x)⇒  x=(√(1+(√e^y )))⇒f^(−1) (x)=(√(1+(√e^x )))
$${e}^{{y}} =\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\sqrt{{e}^{{y}} }=\sqrt{{x}}−\mathrm{1}\Rightarrow\mathrm{1}+\sqrt{{e}^{{y}} }=\sqrt{{x}}\Rightarrow \\ $$$${x}=\sqrt{\mathrm{1}+\sqrt{{e}^{{y}} }}\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt{\mathrm{1}+\sqrt{{e}^{{x}} }} \\ $$
Commented by maxmathsup by imad last updated on 16/Jun/19
1) x∈D_f  ⇔x≥0 and x−2(√x)+1>0 ⇔x≥0 and ((√x)−1)^2 >0 ⇔x≥0 and x≠1 ⇒  D_f =[0,1[∪]1,+∞[  2)f(x)=y ⇔x =f^(−1) (y)  f(x)=y ⇒ln(x+1−2(√x)) =y ⇒x+1−2(√x)=e^y  ⇒((√x)−1)^(2 ) =e^y  ⇒  ∣(√x)−1∣ =(√e^y )    by e^y >0 ⇒ x>1 ⇒(√x)−1 =(√e^y ) ⇒(√x)=1+e^(y/2)  ⇒  x =(1+e^(y/2) )^2   ⇒  f^(−1) (x) =(1+e^(x/2) )^2   3) let I =∫ f(x)dx ⇒ I =∫  ln(x+1−2(√x))dx  by parts  I =xln(x+1−2(√x)) −∫ x((1−(1/( (√x))))/(x+1−2(√x))) dx  =xln(x+1−2(√x)) −∫ x(((√x)−1)/( (√x)(x+1−2(√x)))) dx  ∫  x(((√x)−1)/( (√x)(x+1−2(√x))))dx = ∫  (((√x)((√x)−1))/(x+1−2(√x))) dx = ∫ (((√x)((√x)−1))/(((√x)−1)^2 )) dx =∫  ((√x)/( (√x)−1)) dx  =_((√x)=t)         ∫   (t/(t−1))(2t)dt =2 ∫ ((t^2 −1+1)/(t−1)) dt =2 ∫(t+1)dt +2 ∫ (dt/(t−1))  =2(t^2 /2) +2t +2ln∣t−1∣ +c =t^2  +2t +2ln∣t−1∣ +c  =x +2(√x) +2ln∣(√x)−1∣ +c ⇒  I =xln(x+1−2(√x))−x−2(√x)−2ln∣(√x)−1∣+c .
$$\left.\mathrm{1}\right)\:{x}\in{D}_{{f}} \:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:{x}−\mathrm{2}\sqrt{{x}}+\mathrm{1}>\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:{x}\neq\mathrm{1}\:\Rightarrow \\ $$$${D}_{{f}} =\left[\mathrm{0},\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)={y}\:\Leftrightarrow{x}\:={f}^{−\mathrm{1}} \left({y}\right) \\ $$$${f}\left({x}\right)={y}\:\Rightarrow{ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:={y}\:\Rightarrow{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}={e}^{{y}} \:\Rightarrow\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}\:} ={e}^{{y}} \:\Rightarrow \\ $$$$\mid\sqrt{{x}}−\mathrm{1}\mid\:=\sqrt{{e}^{{y}} }\:\:\:\:{by}\:{e}^{{y}} >\mathrm{0}\:\Rightarrow\:{x}>\mathrm{1}\:\Rightarrow\sqrt{{x}}−\mathrm{1}\:=\sqrt{{e}^{{y}} }\:\Rightarrow\sqrt{{x}}=\mathrm{1}+{e}^{\frac{{y}}{\mathrm{2}}} \:\Rightarrow \\ $$$${x}\:=\left(\mathrm{1}+{e}^{\frac{{y}}{\mathrm{2}}} \right)^{\mathrm{2}} \:\:\Rightarrow\:\:{f}^{−\mathrm{1}} \left({x}\right)\:=\left(\mathrm{1}+{e}^{\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{I}\:=\int\:{f}\left({x}\right){dx}\:\Rightarrow\:{I}\:=\int\:\:{ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right){dx}\:\:{by}\:{parts} \\ $$$${I}\:={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:−\int\:{x}\frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{x}}}}{{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}}\:{dx} \\ $$$$={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:−\int\:{x}\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)}\:{dx} \\ $$$$\int\:\:{x}\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)}{dx}\:=\:\int\:\:\frac{\sqrt{{x}}\left(\sqrt{{x}}−\mathrm{1}\right)}{{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}}\:{dx}\:=\:\int\:\frac{\sqrt{{x}}\left(\sqrt{{x}}−\mathrm{1}\right)}{\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=\int\:\:\frac{\sqrt{{x}}}{\:\sqrt{{x}}−\mathrm{1}}\:{dx} \\ $$$$=_{\sqrt{{x}}={t}} \:\:\:\:\:\:\:\:\int\:\:\:\frac{{t}}{{t}−\mathrm{1}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\frac{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{t}−\mathrm{1}}\:{dt}\:=\mathrm{2}\:\int\left({t}+\mathrm{1}\right){dt}\:+\mathrm{2}\:\int\:\frac{{dt}}{{t}−\mathrm{1}} \\ $$$$=\mathrm{2}\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{2}{t}\:+\mathrm{2}{ln}\mid{t}−\mathrm{1}\mid\:+{c}\:={t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}{ln}\mid{t}−\mathrm{1}\mid\:+{c} \\ $$$$={x}\:+\mathrm{2}\sqrt{{x}}\:+\mathrm{2}{ln}\mid\sqrt{{x}}−\mathrm{1}\mid\:+{c}\:\Rightarrow \\ $$$${I}\:={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)−{x}−\mathrm{2}\sqrt{{x}}−\mathrm{2}{ln}\mid\sqrt{{x}}−\mathrm{1}\mid+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Jun/19
∫ f^(−1) (x)dx =∫(1+e^(x/2) )^2 dx =∫   (1+2 e^(x/2)  +e^x )dx  =x +4e^(x/2)  +e^x  +c .
$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\int\left(\mathrm{1}+{e}^{\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} {dx}\:=\int\:\:\:\left(\mathrm{1}+\mathrm{2}\:{e}^{\frac{{x}}{\mathrm{2}}} \:+{e}^{{x}} \right){dx} \\ $$$$={x}\:+\mathrm{4}{e}^{\frac{{x}}{\mathrm{2}}} \:+{e}^{{x}} \:+{c}\:. \\ $$

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