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Question Number 81043 by john santu last updated on 09/Feb/20

∫ ((2e^(2x) −e^x )/(√(3e^(2x) −6e^x −1))) dx

$$\int\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{dx} \\ $$

Commented by john santu last updated on 09/Feb/20

u = e^x ⇒du = e^x dx ∫ ((2u−1)/(√(3(u−1)^2 −4))) du = ∫ ((2u−1)/(2(√((3/4)(u−1)^2 −1)))) du = (3/4)(u−1)^2 =sec^2 θ ⇒ ∫ ( ((2.(2/(√3))sec θ+1)/(2(√(sec^2 θ−1)))))(2/(√3))sec θtan θ dθ = ∫ (4/3)sec^2 θ dθ +∫ (1/(√3)) sec θ dθ = (4/3)tan θ +(1/(√3)) ln∣sec θ+tan θ∣+c (4/3)(√((3/4)(e^x −1)^2 −1))+(1/(√3)) ln∣((√3)/2)(e^x −1)+(√((3/4)(e^x −1)^2 −1))∣ +c

$${u}\:=\:{e}^{{x}} \:\Rightarrow{du}\:=\:{e}^{{x}} \:{dx} \\ $$$$\int\:\frac{\mathrm{2}{u}−\mathrm{1}}{\sqrt{\mathrm{3}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}\:{du}\:= \\ $$$$\int\:\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}}\:{du}\:= \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\left({u}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{sec}^{\mathrm{2}} \:\theta\: \\ $$$$\Rightarrow\:\int\:\left(\:\frac{\mathrm{2}.\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{sec}\:\theta+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}}}\right)\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{sec}\:\theta\mathrm{tan}\:\theta\:{d}\theta\:= \\ $$$$\int\:\frac{\mathrm{4}}{\mathrm{3}}\mathrm{sec}\:^{\mathrm{2}} \theta\:{d}\theta\:+\int\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{sec}\:\theta\:{d}\theta\:= \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\mathrm{tan}\:\theta\:+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{ln}\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid+{c} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{ln}\mid\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({e}^{{x}} −\mathrm{1}\right)+\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\mid\:+{c} \\ $$

Commented by john santu last updated on 09/Feb/20

maybe someone knows another way?

$${maybe}\:{someone}\:{knows}\:{another}\:{way}? \\ $$

Commented by peter frank last updated on 09/Feb/20

sir the first line 2u^2 −u

$${sir}\:{the}\:{first}\:{line}\:\mathrm{2}{u}^{\mathrm{2}} −{u} \\ $$

Commented by john santu last updated on 09/Feb/20

the same sir to 2u−1

$${the}\:{same}\:{sir}\:{to}\:\mathrm{2}{u}−\mathrm{1} \\ $$

Commented by abdomathmax last updated on 09/Feb/20

let I=∫ ((2e^(2x) −e^x )/(√(3e^(2x) −6e^x −1)))dx changement e^x =t?give I=∫ ((2t^2 −t)/(√(3t^2 −6t−1)))×(dt/t) =∫ ((2t−1)/(√(3t^2 −6t−1)))dt 3t^2 −6t−1=0 →Δ^′ =(−3)^2 +3=12 ⇒ t_1 =((3+2(√3))/3) =1+(2/(√3)) and t_2 =1−(2/(√3)) ⇒ I =∫ ((2t−1)/(√(3(t−t_1 )(t−t_2 ))))dt =(1/(√3))∫ ((2t−1)/((√(t−t_1 ))×(√(t−t_2 ))))dt also changement (√(t−t_1 ))=u give t−t_1 =u^2 and (√3)I=∫ ((2(u^2 +t_1 )−1)/(u(√(u^2 +t_1 −t_2 ))))×(2u)du =2∫ ((2u^2 +2t_1 −1)/(√(u^2 +(4/(√3)))))du =_(u=(2/((^ (√(√3)))))sh(z)) =2 ∫ ((2((4/(√3)))sh^2 (z)+2t_1 −1)/((2/(((√((√)3)))))ch(z)))×(2/(((√((√3))))))ch(z)dz =2 ∫ {(8/(√3))sh^2 (z)+2t_1 −1}dz =((16)/(√3))∫ sh^2 (z)dz +(4t_1 −2)z =((16)/(√3))∫ ((ch(2z)−1)/2)dz +(4t_1 −2)z =(8/(√3)) ∫ ch(2z)dz−(8/(√3))z +(4t_1 −2)z =(4/(√3))sh(2z) +(4t_1 −(8/(√3))−2)z +c z=argsh(((√(√3))/2)u)=ln(((√(√3))/2)u +(√(1+((√3)/4)u^2 ))) and u=(√(t−t_1 ))

$${let}\:{I}=\int\:\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}{dx}\:{changement}\:{e}^{{x}} ={t}?{give} \\ $$$${I}=\int\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} −{t}}{\sqrt{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}}}×\frac{{dt}}{{t}}\:=\int\:\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}}}{dt} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{1}=\mathrm{0}\:\rightarrow\Delta^{'} =\left(−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}=\mathrm{12}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:=\mathrm{1}+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\:{and}\:{t}_{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}}{dt}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\int\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{{t}−{t}_{\mathrm{1}} }×\sqrt{{t}−{t}_{\mathrm{2}} }}{dt} \\ $$$${also}\:{changement}\:\sqrt{{t}−{t}_{\mathrm{1}} }={u}\:{give}\:{t}−{t}_{\mathrm{1}} ={u}^{\mathrm{2}} \:{and} \\ $$$$\sqrt{\mathrm{3}}{I}=\int\:\:\frac{\mathrm{2}\left({u}^{\mathrm{2}} +{t}_{\mathrm{1}} \right)−\mathrm{1}}{{u}\sqrt{{u}^{\mathrm{2}} +{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}×\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}}{\sqrt{{u}^{\mathrm{2}} +\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}}}{du}\:\:=_{{u}=\frac{\mathrm{2}}{\left(^{} \sqrt{\sqrt{\mathrm{3}}}\right)}{sh}\left({z}\right)} \\ $$$$=\mathrm{2}\:\int\:\:\frac{\mathrm{2}\left(\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\right){sh}^{\mathrm{2}} \left({z}\right)+\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}}{\frac{\mathrm{2}}{\left(\sqrt{\sqrt{}\mathrm{3}}\right)}{ch}\left({z}\right)}×\frac{\mathrm{2}}{\left(\sqrt{\left.\sqrt{\mathrm{3}}\right)}\right.}{ch}\left({z}\right){dz} \\ $$$$=\mathrm{2}\:\int\:\left\{\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}{sh}^{\mathrm{2}} \left({z}\right)+\mathrm{2}{t}_{\mathrm{1}} −\mathrm{1}\right\}{dz} \\ $$$$=\frac{\mathrm{16}}{\sqrt{\mathrm{3}}}\int\:{sh}^{\mathrm{2}} \left({z}\right){dz}\:\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z}\: \\ $$$$=\frac{\mathrm{16}}{\sqrt{\mathrm{3}}}\int\:\frac{{ch}\left(\mathrm{2}{z}\right)−\mathrm{1}}{\mathrm{2}}{dz}\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z} \\ $$$$=\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}\:\int\:{ch}\left(\mathrm{2}{z}\right){dz}−\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}{z}\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\mathrm{2}\right){z} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}{sh}\left(\mathrm{2}{z}\right)\:+\left(\mathrm{4}{t}_{\mathrm{1}} −\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}−\mathrm{2}\right){z}\:+{c} \\ $$$${z}={argsh}\left(\frac{\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}}{u}\right)={ln}\left(\frac{\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}}{u}\:+\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{u}^{\mathrm{2}} }\right) \\ $$$${and}\:{u}=\sqrt{{t}−{t}_{\mathrm{1}} } \\ $$

Commented by jagoll last updated on 09/Feb/20

what is argsh sir? sinh^(−1) ?

$${what}\:{is}\:{argsh}\:{sir}? \\ $$$$\mathrm{sinh}^{−\mathrm{1}} \:?\: \\ $$

Commented by Rio Michael last updated on 09/Feb/20

sure

$$\mathrm{sure} \\ $$

Commented by msup trace by abdo last updated on 09/Feb/20

yes

$${yes} \\ $$

Answered by Rio Michael last updated on 09/Feb/20

let u = e^(x ) ⇒ du = e^x dx ∫((2e^(2x) −e^x )/(√(3e^(2x) −6e^x −1))) dx = ∫((2e^x −1)/(√(3e^(2x) −6e^x −1))) e^x dx = ∫((2u−1)/(√(3u^2 −6u−1))) du = ∫((2u)/(√(3u^2 −6u−1)))du(please help me evaluate this) − ∫(1/(√(3u^2 −6u−1))) du = −∫(1/(√(3(u−1)^2 −4))) du =− (1/(√3)) arcosh(((u−1)/2)) + k

$$\mathrm{let}\:{u}\:=\:{e}^{{x}\:} \:\Rightarrow\:{du}\:=\:{e}^{{x}} {dx} \\ $$$$\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{dx}\:=\:\int\frac{\mathrm{2}{e}^{{x}} −\mathrm{1}}{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{e}^{{x}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\frac{\mathrm{2}{u}−\mathrm{1}}{\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}\:{du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\frac{\mathrm{2}{u}}{\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}{du}\left({please}\:{help}\:{me}\:{evaluate}\:{this}\right) \\ $$$$−\:\int\frac{\mathrm{1}}{\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{1}}}\:{du}\:=\:−\int\frac{\mathrm{1}}{\sqrt{\mathrm{3}\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}\:{du}\:=−\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{arcosh}\left(\frac{{u}−\mathrm{1}}{\mathrm{2}}\right)\:+\:{k} \\ $$$$ \\ $$$$ \\ $$

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