Question Number 58299 by maxmathsup by imad last updated on 21/Apr/19
$${find}\:\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Commented by maxmathsup by imad last updated on 24/Apr/19
$${let}\:{I}\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\Rightarrow\:{I}\:=\int\:\:\frac{{dx}}{\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)} \:\:\:\int\:\:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}+{sh}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\int\:\:\frac{{ch}\left({t}\right)}{{ch}^{\mathrm{3}} {t}}\:{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\frac{\mathrm{4}}{\mathrm{3}}}\int\:\:\frac{{dt}}{{ch}^{\mathrm{2}} {t}}\:=\frac{\mathrm{4}}{\mathrm{3}}\:\:\int\:\:\:\frac{{dt}}{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{\mathrm{1}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{\mathrm{2}+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }\:=_{{e}^{\mathrm{2}{t}} ={u}} \:\:\:\frac{\mathrm{16}}{\mathrm{3}}\:\int\:\:\:\:\:\frac{{du}}{\mathrm{2}{u}\left(\mathrm{2}+{u}\:+{u}^{−\mathrm{1}} \right)}\:=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\:\frac{{du}}{\mathrm{2}{u}\:+{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\:\int\:\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{8}}{\mathrm{3}\left({u}+\mathrm{1}\right)}\:+{c}\:\:{but}\:\:{u}\:={e}^{\mathrm{2}{t}} \:\:\:{and}\:\:\:{t}\:={argsh}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)\:\Rightarrow{u}\:=\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{8}}{\mathrm{3}\left(\:\:\:\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}\right)}\:+{C}\:. \\ $$
Answered by Smail last updated on 21/Apr/19
$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\left(\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{3}/\mathrm{2}} \int\frac{{dx}}{\left(\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${let}\:\:{tant}=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$${I}=\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{3}/\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int{costdt}=\frac{\mathrm{4}}{\mathrm{3}}{sint}+{C} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{cost}.{tant}+{C}=\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}×\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }}+{C} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+{C} \\ $$