Question Number 140665 by ERA last updated on 11/May/21
$$\int\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{6}} +\mathrm{x}^{\mathrm{3}} +\mathrm{1}}} \\ $$
Answered by MJS_new last updated on 11/May/21
![∫(dx/(x(√(x^6 +x^3 +1))))= [t=(((√3)(2x^3 +1))/( 3)) → dx=((√3)/(6x^2 ))dt] =((2(√3))/3)∫(dt/((3t−(√3))(√(t^2 +1))))= [u=t+(√(t^2 +1)) → dt=((√(t^2 +1))/(t+(√(t^2 +1))))du] =((4(√3))/3)∫(du/((u−(√3))(3u+(√3))))= =(1/3)∫(du/(u−(√3)))−∫(du/(3u+(√3)))= =(1/3)ln (u−(√3)) −(1/3)ln (3u+(√3)) = =(1/3)ln ((u−(√3))/(3u+(√3))) =... =(1/3)ln ∣((x^3 +2−2(√(x^6 +x^3 +1)))/x^3 )∣ +C](https://www.tinkutara.com/question/Q140670.png)
$$\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{6}} +{x}^{\mathrm{3}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}\right)}{\:\mathrm{3}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}{x}^{\mathrm{2}} }{dt}\right] \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{{dt}}{\left(\mathrm{3}{t}−\sqrt{\mathrm{3}}\right)\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{u}={t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dt}=\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{du}\right] \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{{du}}{\left({u}−\sqrt{\mathrm{3}}\right)\left(\mathrm{3}{u}+\sqrt{\mathrm{3}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{du}}{{u}−\sqrt{\mathrm{3}}}−\int\frac{{du}}{\mathrm{3}{u}+\sqrt{\mathrm{3}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({u}−\sqrt{\mathrm{3}}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{3}{u}+\sqrt{\mathrm{3}}\right)\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\frac{{u}−\sqrt{\mathrm{3}}}{\mathrm{3}{u}+\sqrt{\mathrm{3}}}\:=… \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{3}} +\mathrm{2}−\mathrm{2}\sqrt{{x}^{\mathrm{6}} +{x}^{\mathrm{3}} +\mathrm{1}}}{{x}^{\mathrm{3}} }\mid\:+{C} \\ $$
Commented by ERA last updated on 11/May/21
$$\boldsymbol{\mathrm{t}}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{3}}\:\:\:? \\ $$
Answered by Ar Brandon last updated on 07/Jun/21
$$\mathrm{I}=\int\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{6}} +\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}=\pm\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{6}} }}},\:\mathrm{u}=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\:\:=\mp\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{du}}{\:\sqrt{\mathrm{1}+\mathrm{u}+\mathrm{u}^{\mathrm{2}} }}=\mp\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{du}}{\:\sqrt{\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$\:\:=\mp\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsinh}\left(\frac{\mathrm{2u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mp\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid\left(\frac{\mathrm{2u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\sqrt{\left(\frac{\mathrm{2u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}\mid+\mathrm{C} \\ $$