Question Number 81043 by john santu last updated on 09/Feb/20
$$\int\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{dx} \\ $$
Commented by john santu last updated on 09/Feb/20
$${u}\:=\:{e}^{{x}} \:\Rightarrow{du}\:=\:{e}^{{x}} \:{dx} \\ $$$$\int\:\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}\:{du}\:= \\ $$$$\int\:\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}}\:{du}\:= \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\left({u}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{sec}^{\mathrm{2}} \:\theta\: \\ $$$$\Rightarrow\:\int\:\left(\:\frac{\mathrm{2}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{sec}\:\theta+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}}}\right)\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{sec}\:\theta\mathrm{tan}\:\theta\:{d}\theta\:= \\ $$$$\int\:\frac{\mathrm{4}}{\mathrm{3}}\mathrm{sec}\:^{\mathrm{2}} \theta\:{d}\theta\:+\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sec}\:\theta\:{d}\theta\:= \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\mathrm{tan}\:\theta\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ln}\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid+{c} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ln}\mid\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({e}^{{x}} −\mathrm{1}\right)+\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\mid\:+{c} \\ $$
Commented by john santu last updated on 09/Feb/20
$${maybe}\:{someone}\:{knows}\:{another}\:{way}? \\ $$
Commented by peter frank last updated on 09/Feb/20
$${sir}\:{the}\:{first}\:{line}\:\mathrm{2}{u}^{\mathrm{2}} −{u} \\ $$
Commented by john santu last updated on 09/Feb/20
$${the}\:{same}\:{sir}\:{to}\:\mathrm{2}{u}−\mathrm{1} \\ $$
Commented by abdomathmax last updated on 09/Feb/20
$${let}\:{I}=\int\:\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}{dx}\:{changement}\:{e}^{{x}} ={t}?{give} \\ $$$${I}=\int\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} −{t}}{\:\sqrt{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}}}×\frac{{dt}}{{t}}\:=\int\:\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}}}{dt} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}=\mathrm{0}\:\rightarrow\Delta^{'} =\left(−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}=\mathrm{12}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:=\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:{and}\:{t}_{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}}{dt}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{{t}−{t}_{\mathrm{1}} }×\sqrt{{t}−{t}_{\mathrm{2}} }}{dt} \\ $$$${also}\:{changement}\:\sqrt{{t}−{t}_{\mathrm{1}} }={u}\:{give}\:{t}−{t}_{\mathrm{1}} ={u}^{\mathrm{2}} \:{and} \\ $$$$\sqrt{\mathrm{3}}{I}=\int\:\:\frac{\mathrm{2}\left({u}^{\mathrm{2}} +{t}_{\mathrm{1}} \right)−\mathrm{1}}{{u}\sqrt{{u}^{\mathrm{2}} +{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}×\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} +\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}}}{du}\:\:=_{{u}=\frac{\mathrm{2}}{\left(^{} \sqrt{\sqrt{\mathrm{3}}}\right)}{sh}\left({z}\right)} \\ $$$$=\mathrm{2}\:\int\:\:\frac{\mathrm{2}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right){sh}^{\mathrm{2}} \left({z}\right)+\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}}{\frac{\mathrm{2}}{\left(\sqrt{\sqrt{}\mathrm{3}}\right)}{ch}\left({z}\right)}×\frac{\mathrm{2}}{\left(\sqrt{\left.\sqrt{\mathrm{3}}\right)}\right.}{ch}\left({z}\right){dz} \\ $$$$=\mathrm{2}\:\int\:\left\{\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}{sh}^{\mathrm{2}} \left({z}\right)+\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}\right\}{dz} \\ $$$$=\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}}\int\:{sh}^{\mathrm{2}} \left({z}\right){dz}\:\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z}\: \\ $$$$=\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}}\int\:\frac{{ch}\left(\mathrm{2}{z}\right)−\mathrm{1}}{\mathrm{2}}{dz}\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z} \\ $$$$=\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\int\:{ch}\left(\mathrm{2}{z}\right){dz}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}{z}\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{sh}\left(\mathrm{2}{z}\right)\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}−\mathrm{2}\right){z}\:+{c} \\ $$$${z}={argsh}\left(\frac{\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}}{u}\right)={ln}\left(\frac{\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}}{u}\:+\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{u}^{\mathrm{2}} }\right) \\ $$$${and}\:{u}=\sqrt{{t}−{t}_{\mathrm{1}} } \\ $$
Commented by jagoll last updated on 09/Feb/20
$${what}\:{is}\:{argsh}\:{sir}? \\ $$$$\mathrm{sinh}^{−\mathrm{1}} \:?\: \\ $$
Commented by Rio Michael last updated on 09/Feb/20
$$\mathrm{sure} \\ $$
Commented by msup trace by abdo last updated on 09/Feb/20
$${yes} \\ $$
Answered by Rio Michael last updated on 09/Feb/20
$$\mathrm{let}\:{u}\:=\:{e}^{{x}\:} \:\Rightarrow\:{du}\:=\:{e}^{{x}} {dx} \\ $$$$\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{dx}\:=\:\int\frac{\mathrm{2}{e}^{{x}} −\mathrm{1}}{\:\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{e}^{{x}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}\:{du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\frac{\mathrm{2}{u}}{\:\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}{du}\left({please}\:{help}\:{me}\:{evaluate}\:{this}\right) \\ $$$$−\:\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}\:{du}\:=\:−\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}\:{du}\:=−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{arcosh}\left(\frac{{u}−\mathrm{1}}{\mathrm{2}}\right)\:+\:{k} \\ $$$$ \\ $$$$ \\ $$