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


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

$\int \frac{2 e^{2 x} - e^{x}}{\sqrt{3 e^{2 x} - 6 e^{x} - 1}} d x$
Commented by john santu last updated on 09/Feb/20

$u = e^{x} \Rightarrow d u = e^{x} d x$ $\int \frac{2 u - 1}{\sqrt{3 {(u - 1)}^{2} - 4}} d u =$ $\int \frac{2 u - 1}{2 \sqrt{\frac{3}{4} {(u - 1)}^{2} - 1}} d u =$ $\frac{3}{4} {(u - 1)}^{2} = \sec^{2} θ$ $\Rightarrow \int (\frac{2 . \frac{2}{\sqrt{3}} \sec θ + 1}{2 \sqrt{\sec^{2} θ - 1}}) \frac{2}{\sqrt{3}} \sec θ \tan θ d θ =$ $\int \frac{4}{3} \sec^{2} θ d θ + \int \frac{1}{\sqrt{3}} \sec θ d θ =$ $\frac{4}{3} \tan θ + \frac{1}{\sqrt{3}} l n ∣ \sec θ + \tan θ ∣ + c$ $\frac{4}{3} \sqrt{\frac{3}{4} {(e^{x} - 1)}^{2} - 1} + \frac{1}{\sqrt{3}} l n ∣ \frac{\sqrt{3}}{2} (e^{x} - 1) + \sqrt{\frac{3}{4} {(e^{x} - 1)}^{2} - 1} ∣ + c$
Commented by john santu last updated on 09/Feb/20

$m a y b e s o m e o n e k n o w s a n o t h e r w a y ?$
Commented by peter frank last updated on 09/Feb/20

$s i r t h e f i r s t l i n e 2 u^{2} - u$
Commented by john santu last updated on 09/Feb/20

$t h e s a m e s i r t o 2 u - 1$
Commented by abdomathmax last updated on 09/Feb/20
$let I=∫ ((2e^(2x) −e^x )/(√(3e^(2x) −6e^x −1)))dx changement e^x =t?give I=∫ ((2t^2 −t)/(√(3t^2 −6t−1)))×(dt/t) =∫ ((2t−1)/(√(3t^2 −6t−1)))dt 3t^2 −6t−1=0 →Δ^′ =(−3)^2 +3=12 ⇒ t_1 =((3+2(√3))/3) =1+(2/(√3)) and t_2 =1−(2/(√3)) ⇒ I =∫ ((2t−1)/(√(3(t−t_1 )(t−t_2 ))))dt =(1/(√3))∫ ((2t−1)/((√(t−t_1 ))×(√(t−t_2 ))))dt also changement (√(t−t_1 ))=u give t−t_1 =u^2 and (√3)I=∫ ((2(u^2 +t_1 )−1)/(u(√(u^2 +t_1 −t_2 ))))×(2u)du =2∫ ((2u^2 +2t_1 −1)/(√(u^2 +(4/(√3)))))du =_(u=(2/((^ (√(√3)))))sh(z)) =2 ∫ ((2((4/(√3)))sh^2 (z)+2t_1 −1)/((2/(((√((√)3)))))ch(z)))×(2/(((√((√3))))))ch(z)dz =2 ∫ {(8/(√3))sh^2 (z)+2t_1 −1}dz =((16)/(√3))∫ sh^2 (z)dz +(4t_1 −2)z =((16)/(√3))∫ ((ch(2z)−1)/2)dz +(4t_1 −2)z =(8/(√3)) ∫ ch(2z)dz−(8/(√3))z +(4t_1 −2)z =(4/(√3))sh(2z) +(4t_1 −(8/(√3))−2)z +c z=argsh(((√(√3))/2)u)=ln(((√(√3))/2)u +(√(1+((√3)/4)u^2 ))) and u=(√(t−t_1 ))$
$l e t I = \int \frac{2 e^{2 x} - e^{x}}{\sqrt{3 e^{2 x} - 6 e^{x} - 1}} d x c h a n g e m e n t e^{x} = t ? g i v e$ $I = \int \frac{2 t^{2} - t}{\sqrt{3 t^{2} - 6 t - 1}} \times \frac{d t}{t} = \int \frac{2 t - 1}{\sqrt{3 t^{2} - 6 t - 1}} d t$ $3 t^{2} - 6 t - 1 = 0 \to Δ^{^{'}} = {(- 3)}^{2} + 3 = 12 \Rightarrow$ $t_{1} = \frac{3 + 2 \sqrt{3}}{3} = 1 + \frac{2}{\sqrt{3}} a n d t_{2} = 1 - \frac{2}{\sqrt{3}} \Rightarrow$ $I = \int \frac{2 t - 1}{\sqrt{3 (t - t_{1}) (t - t_{2})}} d t = \frac{1}{\sqrt{3}} \int \frac{2 t - 1}{\sqrt{t - t_{1}} \times \sqrt{t - t_{2}}} d t$ $a l s o c h a n g e m e n t \sqrt{t - t_{1}} = u g i v e t - t_{1} = u^{2} a n d$ $\sqrt{3} I = \int \frac{2 (u^{2} + t_{1}) - 1}{u \sqrt{u^{2} + t_{1} - t_{2}}} \times (2 u) d u$ $= 2 \int \frac{2 u^{2} + 2 t_{1} - 1}{\sqrt{u^{2} + \frac{4}{\sqrt{3}}}} d u =_{u = \frac{2}{(^{} \sqrt{\sqrt{3}})} s h (z)}$ $= 2 \int \frac{2 (\frac{4}{\sqrt{3}}) {s h}^{2} (z) + 2 t_{1} - 1}{\frac{2}{(\sqrt{\sqrt{} 3})} c h (z)} \times \frac{2}{(\sqrt{\sqrt{3})}} c h (z) d z$ $= 2 \int {\frac{8}{\sqrt{3}} {s h}^{2} (z) + 2 t_{1} - 1} d z$ $= \frac{16}{\sqrt{3}} \int {s h}^{2} (z) d z + (4 t_{1} - 2) z$ $= \frac{16}{\sqrt{3}} \int \frac{c h (2 z) - 1}{2} d z + (4 t_{1} - 2) z$ $= \frac{8}{\sqrt{3}} \int c h (2 z) d z - \frac{8}{\sqrt{3}} z + (4 t_{1} - 2) z$ $= \frac{4}{\sqrt{3}} s h (2 z) + (4 t_{1} - \frac{8}{\sqrt{3}} - 2) z + c$ $z = a r g s h (\frac{\sqrt{\sqrt{3}}}{2} u) = l n (\frac{\sqrt{\sqrt{3}}}{2} u + \sqrt{1 + \frac{\sqrt{3}}{4} u^{2}})$ $a n d u = \sqrt{t - t_{1}}$
Commented by jagoll last updated on 09/Feb/20

$w h a t i s a r g s h s i r ?$ $\sinh^{- 1} ?$
Commented by Rio Michael last updated on 09/Feb/20

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

$y e s$
	Answered by Rio Michael last updated on 09/Feb/20

	$let u = e^{x} \Rightarrow d u = e^{x} d x$ $\int \frac{2 e^{2 x} - e^{x}}{\sqrt{3 e^{2 x} - 6 e^{x} - 1}} d x = \int \frac{2 e^{x} - 1}{\sqrt{3 e^{2 x} - 6 e^{x} - 1}} e^{x} d x$ $= \int \frac{2 u - 1}{\sqrt{3 u^{2} - 6 u - 1}} d u$ $= \int \frac{2 u}{\sqrt{3 u^{2} - 6 u - 1}} d u (p l e a s e h e l p m e e v a l u a t e t h i s)$ $- \int \frac{1}{\sqrt{3 u^{2} - 6 u - 1}} d u = - \int \frac{1}{\sqrt{3 {(u - 1)}^{2} - 4}} d u = - \frac{1}{\sqrt{3}} a r c o s h (\frac{u - 1}{2}) + k$
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