Question Number 124004 by john_santu last updated on 30/Nov/20
$$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:?\: \\ $$
Answered by liberty last updated on 30/Nov/20
$${T}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:;\:\left[\:{let}\:{u}^{\mathrm{3}} =\mathrm{tan}\:^{\mathrm{2}} {x}\:\wedge\:{dx}\:=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} \:\left({u}^{\mathrm{3}} +\mathrm{1}\right)}{du}\:\right] \\ $$$${T}\:=\:\int\:\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} .{u}^{\mathrm{3}/\mathrm{2}} \left({u}^{\mathrm{3}} +\mathrm{1}\right)}\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{{u}^{\mathrm{3}} +\mathrm{1}} \\ $$$${T}=\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\int\frac{{du}}{{u}+\mathrm{1}}\:−\:\int\:\frac{{u}−\mathrm{2}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}\:\right\} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{u}−\mathrm{1}−\mathrm{3}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}\: \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\mid{u}^{\mathrm{2}} −{u}+\mathrm{1}\mid\:+\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\mid\:+\frac{\mathrm{3}}{\mathrm{2}}.\mathrm{2}\:\mathrm{arctan}\:\left(\frac{{u}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)+{c} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\mid+\mathrm{3}\:\mathrm{arctan}\:\left(\mathrm{2}{u}−\mathrm{1}\right)+{c} \\ $$$$\therefore\:{T}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:\:+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{4}} {x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:+\mathrm{1}\:}\mid\:+\mathrm{3arctan}\:\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:−\mathrm{1}\:\right)+{c}\: \\ $$