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Question Number 58299 by maxmathsup by imad last updated on 21/Apr/19

find ∫ (dx/((x^2 +x+1)^(3/2) ))

$${find}\:\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Commented by maxmathsup by imad last updated on 24/Apr/19

let I =∫ (dx/((x^2 +x +1)^(3/2) )) ⇒ I =∫ (dx/(((x+(1/2))^2 +(3/4))^(3/2) )) =_(x+(1/2) =((√3)/2)sh(t)) ∫ (1/(((3/4))^(3/2) (1+sh^2 t)^(3/2) )) ((√3)/2)ch(t) =((√3)/2) ((4/3))^(3/2) ∫ ((ch(t))/(ch^3 t)) dt =((√3)/2) (4/3)(√(4/3))∫ (dt/(ch^2 t)) =(4/3) ∫ (dt/((1+ch(2t))/2)) =(8/3) ∫ (dt/(1+((e^(2t) +e^(−2t) )/2))) =((16)/3) ∫ (dt/(2+e^(2t) +e^(−2t) )) =_(e^(2t) =u) ((16)/3) ∫ (du/(2u(2+u +u^(−1) ))) =(8/3) ∫ (du/(2u +u^2 +1)) =(8/3) ∫ (du/((u+1)^2 )) =−(8/(3(u+1))) +c but u =e^(2t) and t =argsh(((2x+1)/(√3))) =ln(((2x+1)/(√3)) +(√(1+(((2x+1)/(√3)))^2 ))) ⇒u =(((2x+1)/(√3)) +(√(1+(((2x+1)/(√3)))^2 )))^2 ⇒ I =−(8/(3( (((2x+1)/(√3)) +(√(1+(((2x+1)/(√3)))^2 )))^2 +1))) +C .

$${let}\:{I}\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\Rightarrow\:{I}\:=\int\:\:\frac{{dx}}{\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)} \:\:\:\int\:\:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}+{sh}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\int\:\:\frac{{ch}\left({t}\right)}{{ch}^{\mathrm{3}} {t}}\:{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\frac{\mathrm{4}}{\mathrm{3}}}\int\:\:\frac{{dt}}{{ch}^{\mathrm{2}} {t}}\:=\frac{\mathrm{4}}{\mathrm{3}}\:\:\int\:\:\:\frac{{dt}}{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{\mathrm{1}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{\mathrm{2}+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }\:=_{{e}^{\mathrm{2}{t}} ={u}} \:\:\:\frac{\mathrm{16}}{\mathrm{3}}\:\int\:\:\:\:\:\frac{{du}}{\mathrm{2}{u}\left(\mathrm{2}+{u}\:+{u}^{−\mathrm{1}} \right)}\:=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\:\frac{{du}}{\mathrm{2}{u}\:+{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{8}}{\mathrm{3}\left({u}+\mathrm{1}\right)}\:+{c}\:\:{but}\:\:{u}\:={e}^{\mathrm{2}{t}} \:\:\:{and}\:\:\:{t}\:={argsh}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)\:\Rightarrow{u}\:=\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{8}}{\mathrm{3}\left(\:\:\:\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}\right)}\:+{C}\:. \\ $$

Answered by Smail last updated on 21/Apr/19

x^2 +x+1=(x+(1/2))^2 +(3/4)=(3/4)((((2x+1)/(√3)))^2 +1) I=∫(dx/((x^2 +x+1)^(3/2) ))=((4/3))^(3/2) ∫(dx/(((((2x+1)/(√3)))^2 +1)^(3/2) )) let tant=((2x+1)/(√3))⇒dx=((√3)/2)(1+tan^2 t)dt I=(4/3)∫(((1+tan^2 t))/((1+tan^2 t)^(3/2) ))dt =(4/3)∫costdt=(4/3)sint+C =(4/3)cost.tant+C=(4/3)×((2x+1)/((√3)×(√(1+(((2x+1)/(√3)))^2 ))))+C ∫(dx/((x^2 +x+1)^(3/2) ))=((2(2x+1))/(3(√(x^2 +x+1))))+C

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\left(\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{3}/\mathrm{2}} \int\frac{{dx}}{\left(\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${let}\:\:{tant}=\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\Rightarrow{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$${I}=\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{3}/\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int{costdt}=\frac{\mathrm{4}}{\mathrm{3}}{sint}+{C} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{cost}.{tant}+{C}=\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}×\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }}+{C} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+{C} \\ $$

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