Question Number 35139 by behi83417@gmail.com last updated on 15/May/18
Commented by abdo mathsup 649 cc last updated on 16/May/18
![let put I = ∫_0 ^(√2) ((x+1)/(√(x^2 +x+1)))dx I =(1/2)∫_0 ^(√2) ((2x+2)/(√(x^2 +x+1)))dx =(1/2) ∫_0 ^(√2) ((2x +1)/(√(x^2 +x+1)))dx +(1/2) ∫_0 ^(√2) (dx/(√(x^2 +x+1))) but ∫_0 ^(√2) ((2x+1)/(√(x^2 +x+1)))dx =[ 2(√(x^2 +x+1))]_0 ^(√2) =2(√(3+(√2))) −2 we have x^2 +x+1=(x+(1/2))^2 +(3/4) and changement x+(1/2) = ((√3)/2) sh(t) give sh(t)=((2x+1)/(√3)) ∫_0 ^(√2) (dx/(√(x^2 +x+1))) = ∫_(argsh((1/(√3)))) ^(argsh(((2(√2) +1)/(√3)))) (1/(√( (3/4)( 1+sh^2 t)))) ((√3)/2)chtdt = ∫_(argsh((1/(√3)))) ^(argsh(((2(√2) +1)/(√3)))) dt = argsh(((2(√2) +1)/(√3))) − argsh((1/(√3))) =ln( ((2(√2) +1)/(√3)) +(√(1+(((2(√2) +1)/(√3)))^2 )) ) −ln ( (1/(√3)) +(√(1+((1/(√3)))^2 )) ) so I =(√(3+(√2))) −1 +(1/2)ln( ((2(√2) +1)/(√3)) +(√(1+(((2(√2) +1)/(√3)))^2 ))) −ln((1/(√3)) + (√(1+((1/(√3)))^2 ) .))](Q35146.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}{dx} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{2}{x}+\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{2}{x}\:+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}{dx}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}{dx}\:=\left[\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}\:\:−\mathrm{2}\:\:{we}\:{have}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${and}\:{changement}\:\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{sh}\left({t}\right)\:{give}\:{sh}\left({t}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}\:=\:\int_{{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\:\frac{\mathrm{3}}{\mathrm{4}}\left(\:\mathrm{1}+{sh}^{\mathrm{2}} {t}\right)}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{chtdt} \\ $$$$=\:\int_{{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \:\:\:\:{dt}\:=\:{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:−\:{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\ $$$$={ln}\left(\:\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\:\right) \\ $$$$−{ln}\:\left(\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\:\right)\:{so} \\ $$$${I}\:=\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}\:−\mathrm{1}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right) \\ $$$$−{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\:\sqrt{\left.\mathrm{1}+\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \:\right)\:.}\right. \\ $$
Commented by abdo mathsup 649 cc last updated on 16/May/18
![let put x+1=t ⇒ I = ∫_1 ^(1+(√2)) (t/(√((t−1)^2 +t)))dt = ∫_1 ^(1+(√2)) (t/(√(t^2 −t +1))) dt = ∫_1 ^(1+(√2)) (t/(√((t−(1/2))^2 +(3/4))))dt changement (t−(1/2))^ =((√3)/2) u ⇒ u=((2t−1)/(√3)) I = ∫_(1/(√3)) ^((1+2(√2))/(√3)) ((1 +u(√3))/(2((√3)/2)(√(u^2 +1))))du I = (1/(√3)) ∫_(1/(√3)) ^((1+2(√2))/(√3)) (du/(√(1+u^2 ))) + ∫_(1/(√3)) ^((1+2(√2))/(√3)) (u/(√(u^2 +1)))du I =(1/(√3))[ln(u +(√(1+u^2 ))]_(1/(√3)) ^((1+2(√2))/(√3)) +[(√(1+u^2 ))]_(1/(√3)) ^((1+2(√2))/(√3)) so the value of I is determined....](Q35209.png)
$${let}\:{put}\:{x}+\mathrm{1}={t}\:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\:\:\:\:\frac{{t}}{\sqrt{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+{t}}}{dt} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\frac{{t}}{\sqrt{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}}\:{dt} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\:\:\frac{{t}}{\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}}{dt}\:\:{changement} \\ $$$$\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}\:\Rightarrow\:{u}=\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${I}\:=\:\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}} \:\:\:\:\:\frac{\mathrm{1}\:+{u}\sqrt{\mathrm{3}}}{\mathrm{2}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{{u}^{\mathrm{2}} \:+\mathrm{1}}}{du} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}} \:\:\:\:\frac{{du}}{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:\:+\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}} \:\:\:\:\:\frac{{u}}{\sqrt{{u}^{\mathrm{2}} \:+\mathrm{1}}}{du} \\ $$$${I}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left[{ln}\left({u}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}} \:\:+\left[\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}} \:{so}\right. \\ $$$${the}\:{value}\:{of}\:{I}\:{is}\:{determined}.... \\ $$
Answered by MJS last updated on 16/May/18
![∫((x+1)/(√(x^2 +x+1)))dx=∫(((1/2)(2x+1)+(1/2))/(√(x^2 +x+1)))dx= =(1/2)∫((2x+1)/(√(x^2 +x+1)))dx+(1/2)∫(dx/(√(x^2 +x+1)))= (1/2)∫((2x+1)/(√(x^2 +x+1)))dx= [t=x^2 +x+1 → dx=(dt/(2x+1))] =(1/2)∫(dt/(√t))=(1/2)2(√t)=(√(x^2 +x+1)) (1/2)∫(dx/(√(x^2 +x+1)))=(1/2)∫(dx/(√((x+(1/2))^2 +(3/4))))= [u=((√3)/3)(2x+1) → dx=((√3)/2)du] =(1/2)∫(du/(√(u^2 +1)))=(1/2)ln∣(√(u^2 +1))+u∣= =(1/2)ln∣(√((((√3)/3)(2x+1))^2 +1))+((√3)/2)(2x+1)∣= =(1/2)ln∣2(√(x^2 +x+1))+2x+1∣−((ln(3))/4) =(√(x^2 +x+1))+(1/2)ln∣2(√(x^2 +x+1))+2x+1∣+C ∫_0 ^(√2) ((x+1)/(√(x^2 +x+1)))dx=(√(3+(√2)))+(1/2)ln(2(√(3+(√2)))+2(√2)+1)−(1+((ln(3))/2))= =(√(3+(√2)))−1+(1/2)ln(((2(√(3+(√2)))+2(√2)+1)/3))≈1.593316](Q35148.png)
$$\int\frac{{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}=\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} +{x}+\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}+\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\sqrt{{t}}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}\sqrt{{t}}=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}+{u}\mid= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)\right)^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{1}\right)\mid= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\mathrm{2}{x}+\mathrm{1}\mid−\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{4}} \\ $$$$ \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\mathrm{2}{x}+\mathrm{1}\mid+{C} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}=\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)−\left(\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{2}}\right)= \\ $$$$=\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{3}}\right)\approx\mathrm{1}.\mathrm{593316} \\ $$