2e-2x-e-x-3e-2x-6e-x-1-dx-

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

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