Question Number 167330 by LEKOUMA last updated on 13/Mar/22
$${Calculate} \\ $$$$\int\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$
Answered by greogoury55 last updated on 13/Mar/22
![=∫ (((√(x^2 +x+1))−x)/(x+1)) dx =∫ (((√((x+(1/2))^2 +(3/4)))−x)/(x+1)) dx K_1 = ∫ ((√((x+(1/2))^2 +(3/4)))/(x+1)) dx [ x+(1/2)=((√3)/2) tan t ] K_1 =∫ ((((√3)/2) sec t )/((1/2)(1+(√3) tan t))) ((√3)/2).sec^2 t dt K_1 = (3/2)∫ ((sec^2 t)/(cos t+(√3) sin t)) dt K_1 =(3/2) ∫ ((sec^2 t)/(2((1/2)cos t+((√3)/2) sin t))) dt K_1 =(3/4)∫ ((sec^2 t)/(cos (t−(π/3)))) dt K_2 =∫ (x/(x+1))dx=∫ (((x+1)−1)/(x+1)) dx K_2 =x−ln ∣x+1∣ +c](Q167340.png)
$$=\int\:\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−{x}}{{x}+\mathrm{1}}\:{dx}\: \\ $$$$=\int\:\frac{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}−{x}}{{x}+\mathrm{1}}\:{dx} \\ $$$${K}_{\mathrm{1}} =\:\int\:\frac{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}{{x}+\mathrm{1}}\:{dx} \\ $$$$\:\left[\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{tan}\:{t}\:\right] \\ $$$${K}_{\mathrm{1}} =\int\:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sec}\:{t}\:}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:{t}\right)}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\mathrm{sec}^{\mathrm{2}} {t}\:{dt} \\ $$$${K}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}}{\mathrm{cos}\:{t}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:{t}}\:{dt} \\ $$$${K}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:{t}\right)}\:{dt} \\ $$$${K}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}}\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}}{\mathrm{cos}\:\left({t}−\frac{\pi}{\mathrm{3}}\right)}\:{dt} \\ $$$${K}_{\mathrm{2}} =\int\:\frac{{x}}{{x}+\mathrm{1}}{dx}=\int\:\frac{\left({x}+\mathrm{1}\right)−\mathrm{1}}{{x}+\mathrm{1}}\:{dx} \\ $$$${K}_{\mathrm{2}} ={x}−\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:+{c}\:\: \\ $$
Commented by peter frank last updated on 14/Mar/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by MJS_new last updated on 13/Mar/22
![∫(dx/(x+(√(x^2 +x+1))))=−∫((x−(√(x^2 +x+1)))/(x+1))dx= [t=x+(1/2) → dx=dt] =−∫((2t−1−(√(4t^2 +3)))/(2t+1))dt= [u=((√3)/3)(2t+(√(4t^2 +3))) → dt=(((√3)(u^2 +1))/(4u^2 ))] =((√3)/2)∫((u^2 +1)/(u^2 ((√3)u−1)))du= =∫(((2(√3))/( (√3)u−1))−(3/(2u))−((√3)/(2u^2 )))du= =2ln ((√3)u−1) −(3/2)ln u +((√3)/(2u))= ... =2ln (x+(√(x^2 +x+1))) −(3/2)ln (2x+1+(√(x^2 +x+1))) −x+(√(x^2 +x+1))+C](Q167342.png)
$$\int\frac{{dx}}{{x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}=−\int\frac{{x}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=−\int\frac{\mathrm{2}{t}−\mathrm{1}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{2}{t}+\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}}\right)\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{3}}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}{u}^{\mathrm{2}} }\right] \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}{u}−\mathrm{1}\right)}{du}= \\ $$$$=\int\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}{u}−\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{u}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{u}^{\mathrm{2}} }\right){du}= \\ $$$$=\mathrm{2ln}\:\left(\sqrt{\mathrm{3}}{u}−\mathrm{1}\right)\:−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:{u}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{u}}= \\ $$$$... \\ $$$$=\mathrm{2ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)\:−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)\:−{x}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+{C} \\ $$