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Question Number 47851 by maxmathsup by imad last updated on 15/Nov/18
let f(x)=x+1+(√x) and g(x)=x+1−(√x) find ∫ ((f(x))/(g(x)))dx and ((∫f(x)dx)/(∫g(x)dx)) .
$${let}\:\:{f}\left({x}\right)={x}+\mathrm{1}+\sqrt{{x}}\:{and}\:{g}\left({x}\right)={x}+\mathrm{1}−\sqrt{{x}} \\ $$$${find}\:\int\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}{dx}\:\:{and}\:\:\frac{\int{f}\left({x}\right){dx}}{\int{g}\left({x}\right){dx}}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Nov/18
let I =∫ ((f(x))/(g(x)))dx ⇒I =∫((x+1+(√x))/(x+1−(√x)))dx cha7gement (√x)=t give I =∫ ((t^2 +t+1)/(t^2 −t +1)) (2t)dt =2 ∫ ((t^3 +t^2 +1)/(t^2 −t +1))dt =2 ∫ ((t(t^2 −t+1)+t^2 −t+t^2 +1)/(t^2 −t +1))dt =2 ∫ t dt +2 ∫ ((2t^2 −t+1)/(t^2 −t +1))dt =t^2 +2 ∫ ((2(t^2 −t+1)+2t−2−t+1)/(t^2 −t+1))dt =t^2 +4t + ∫ ((2t−1−1)/(t^2 −t +1))dt =t^2 +4t +ln∣t^2 −t+1∣−∫ (dt/(t^2 −t +1)) but ∫ (dt/(t^2 −t +1)) =∫ (dt/(t^2 −2(t/2) +(1/(4 )) +(3/4))) =∫ (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u ) (4/3) ∫ (1/(1+u^2 )) ((√3)/2)du =(2/( (√3))) arctan(((2t−1)/( (√3)))) ⇒ I =t^2 +4t −(2/( (√3))) arctan(((2t−1)/( (√3))))+C .
$${let}\:{I}\:=\int\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}{dx}\:\Rightarrow{I}\:=\int\frac{{x}+\mathrm{1}+\sqrt{{x}}}{{x}+\mathrm{1}−\sqrt{{x}}}{dx}\:{cha}\mathrm{7}{gement}\:\sqrt{{x}}={t}\:{give} \\ $$$${I}\:=\int\:\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\:\:\:\frac{{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}{dt} \\ $$$$=\mathrm{2}\:\int\:\frac{{t}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)+{t}^{\mathrm{2}} −{t}+{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt}\:=\mathrm{2}\:\int\:{t}\:{dt}\:+\mathrm{2}\:\int\:\frac{\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt} \\ $$$$={t}^{\mathrm{2}} \:+\mathrm{2}\:\int\:\:\frac{\mathrm{2}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)+\mathrm{2}{t}−\mathrm{2}−{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:\:+\:\int\:\frac{\mathrm{2}{t}−\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+{ln}\mid{t}^{\mathrm{2}} −{t}+\mathrm{1}\mid−\int\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}\:{but} \\ $$$$\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}\:=\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}\:}\:+\frac{\mathrm{3}}{\mathrm{4}}}\:=\int\:\:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}\:} \:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$${I}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Nov/18
but t=(√x) ⇒ I =x +4(√x) −(2/( (√3))) arctan(((2(√x)−1)/( (√3)))) +C .
$${but}\:{t}=\sqrt{{x}}\:\Rightarrow\:{I}\:={x}\:+\mathrm{4}\sqrt{{x}}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}\sqrt{{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{C}\:. \\ $$
Commented by maxmathsup by imad last updated on 18/Nov/18
changement (√x)=t give ∫ f(x)dx=∫ (t^2 +1+t)(2t)dt =2 ∫ (t^3 +t^2 +t)dt =(1/2)t^4 +(2/3)t^3 +t^2 +c_1 =(x^2 /2) +(2/3)x(√x)+x +c_1 ∫ g(x)dx =∫ (x+1−(√x))dx =_((√x)=t) ∫ (t^2 +1−t)(2t)dt =2 ∫ (t^3 +t −t^2 )dt =(t^4 /4) −(2/3)t^3 +t^2 +c_2 =(x^2 /2) −(2/3)x(√x)+x +c_2 ⇒ ((∫f(x)dx)/(∫g(x)dx)) =(((x^2 /2) +(2/3)x(√x)+x +c_1 )/((x^2 /2)−(2/3)x(√x)+x +c_2 )) .
$${changement}\:\sqrt{{x}}={t}\:{give}\:\int\:{f}\left({x}\right){dx}=\int\:\left({t}^{\mathrm{2}} \:+\mathrm{1}+{t}\right)\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:\left({t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:+{t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{4}} \:+\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:\:+{t}^{\mathrm{2}} \:+{c}_{\mathrm{1}} =\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}{x}\sqrt{{x}}+{x}\:+{c}_{\mathrm{1}} \\ $$$$\int\:{g}\left({x}\right){dx}\:=\int\:\left({x}+\mathrm{1}−\sqrt{{x}}\right){dx}\:=_{\sqrt{{x}}={t}} \:\:\:\int\:\left({t}^{\mathrm{2}} \:+\mathrm{1}−{t}\right)\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:\:\left({t}^{\mathrm{3}} \:+{t}\:−{t}^{\mathrm{2}} \right){dt}\:=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:+{c}_{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\mathrm{2}}{\mathrm{3}}{x}\sqrt{{x}}+{x}\:+{c}_{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\int{f}\left({x}\right){dx}}{\int{g}\left({x}\right){dx}}\:=\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}{x}\sqrt{{x}}+{x}\:+{c}_{\mathrm{1}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}{x}\sqrt{{x}}+{x}\:+{c}_{\mathrm{2}} }\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18
∫f(x)dx ∫x+1+(√x) dx =(x^2 /2)+x+(x^(3/2) /(3/2))+c_1 ∫g(x)dx=(x^2 /2)+x−(x^(3/2) /(3/2))+c_2 so ((∫f(x)dx)/(∫g(x)dx))=(((x^2 /2)+x+(x^(3/2) /(3/2))+c_1 )/((x^2 /2)+x−(x^(3/2) /(3/2))+c_2 ))
$$\int{f}\left({x}\right){dx} \\ $$$$\int{x}+\mathrm{1}+\sqrt{{x}}\:{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}+\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{1}} \\ $$$$\int{g}\left({x}\right){dx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}−\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{2}} \\ $$$${so}\:\frac{\int{f}\left({x}\right){dx}}{\int{g}\left({x}\right){dx}}=\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}+\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{1}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}−\frac{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{c}_{\mathrm{2}} } \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18
∫((x+1+(√x) )/(x+1−(√x) ))dx ∫(((√x) +(1/( (√x)))+1)/( (√x) +(1/( (√x)))−1))dx ∫(((√x) +(1/( (√x) ))−1+2)/( (√x) +(1/( (√x) ))−1))dx ∫dx+2∫(dx/( (√x)+(1/( (√x)))−1)) t^2 =x dx=2tdt I_1 =∫dx=x+c_1 I_2 =∫((2tdt)/(t+(1/t)−1))dt ∫((2t^2 )/(t^2 +1−t))dt 2∫((t^2 −t+1+t−1)/(t^2 −t+1))dt 2∫dt+2∫((t−1)/(t^2 −t+1))dt 2∫dt+∫((2t−1−1)/(t^2 −t+1))dt 2∫dt+∫((d(t^2 −t+1))/(t^2 −t+1))−∫(dt/((t^2 −2.t.(1/2)+(1/4)+(3/4)))) 2∫dt+∫((d(t^2 −t+1))/(t^2 −t+1))−∫(dt/((t−(1/2))^2 +(((√3)/2))^2 )) 2t−ln(t^2 −t+1)−(2/(((√3) )/2))tan^(−1) (((t−(1/2))/(((√3) )/2)))+c 2(√x) −ln(x−(√x) +1)−(4/( (√3) ))tan^(−1) ((((√x) −(1/2))/(((√3) )/2)))+c
$$\int\frac{{x}+\mathrm{1}+\sqrt{{x}}\:}{{x}+\mathrm{1}−\sqrt{{x}}\:}{dx} \\ $$$$\int\frac{\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\mathrm{1}}{\:\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}}{dx} \\ $$$$\int\frac{\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}−\mathrm{1}+\mathrm{2}}{\:\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}−\mathrm{1}}{dx} \\ $$$$\int{dx}+\mathrm{2}\int\frac{{dx}}{\:\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}} \\ $$$${t}^{\mathrm{2}} ={x}\:\:\:{dx}=\mathrm{2}{tdt} \\ $$$${I}_{\mathrm{1}} =\int{dx}={x}+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{2}{tdt}}{{t}+\frac{\mathrm{1}}{{t}}−\mathrm{1}}{dt} \\ $$$$\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}−{t}}{dt} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{2}} −{t}+\mathrm{1}+{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int{dt}+\mathrm{2}\int\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int{dt}+\int\frac{\mathrm{2}{t}−\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{2}\int{dt}+\int\frac{{d}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}−\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\mathrm{2}\int{dt}+\int\frac{{d}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}−\int\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}{t}−{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)−\frac{\mathrm{2}}{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}}\right)+{c} \\ $$$$\mathrm{2}\sqrt{{x}}\:−{ln}\left({x}−\sqrt{{x}}\:+\mathrm{1}\right)−\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}\:}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}}\right)+{c} \\ $$

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