If-log-2-x-1-e-x-1-dx-pi-6-then-x-

Question Number 64265 by Chi Mes Try last updated on 16/Jul/19

If \underset{\log 2}{\int^{x}} \frac{1}{\sqrt{e^{x} - 1}} d x = \frac{π}{6}, then x =

Commented by mathmax by abdo last updated on 17/Jul/19

l e t I = \int_{l n 2}^{x} \frac{d x}{\sqrt{e^{x} - 1}} c h a n g e m e n t \sqrt{e^{x} - 1} = t g i v e e^{x} - 1 = t^{2} \Rightarrow

e^{x} = 1 + t^{2} \Rightarrow x = l n (1 + t^{2}) \Rightarrow I = \int_{1}^{\sqrt{e^{x} - 1}} \frac{2 t}{(1 + t^{2}) t} d t

= 2 \int_{1}^{\sqrt{e^{x} - 1}} \frac{d t}{1 + t^{2}} = 2 {[a r c t a n (t)]}_{1}^{\sqrt{e^{x} - 1}} = 2 {a r c t a n \sqrt{e^{x} - 1} - \frac{π}{4}}

I = \frac{π}{6} \Rightarrow 2 a r c t a n \sqrt{e^{x} - 1} - \frac{π}{2} = \frac{π}{6} \Rightarrow 2 a r c t a n \sqrt{e^{x} - 1} = \frac{π}{2} + \frac{π}{6} = \frac{2 π}{3}

\Rightarrow \sqrt{e^{x} - 1} = \frac{π}{3} \Rightarrow e^{x} - 1 = \frac{π^{2}}{9} \Rightarrow e^{x} = 1 + \frac{π^{2}}{9} \Rightarrow x = l n (1 + \frac{π^{2}}{9}) .

Commented by mathmax by abdo last updated on 17/Jul/19

e r r o r a t f i n a l l i n e a r c t a n \sqrt{e^{x} - 1} = \frac{π}{3} \Rightarrow \sqrt{e^{x} - 1} = t a n (\frac{π}{3}) = \sqrt{3} \Rightarrow

e^{x} - 1 = 3 \Rightarrow e^{x} = 4 \Rightarrow x = l n (4) = 2 l n (2) .

Answered by Tanmay chaudhury last updated on 16/Jul/19

\int \frac{d x}{\sqrt{e^{x} - 1}} t^{2} = e^{x} - 1 2 t \times \frac{d t}{d x} = e^{x}

\frac{2 t}{1 + t^{2}} d t = d x

\int \frac{2 t}{1 + t^{2}} \times \frac{1}{t} d t

2 {t a n}^{- 1} (t) + c

2 {t a n}^{- 1} (\sqrt{e^{x} - 1}) + c

\int_{0}^{x} \frac{d x}{\sqrt{e^{x} - 1}}

= 2 ∣ {t a n}^{- 1} \sqrt{e^{x} - 1} ∣_{l n 2}^{x}

= 2 [({t a n}^{- 1} \sqrt{e^{x} - 1}) - {t a n}^{- 1} (\sqrt{e^{l n 2} - 1})]

= 2 [{t a n}^{- 1} \sqrt{e^{x} - 1} - \frac{π}{4}]

s o {t a n}^{- 1} \sqrt{e^{x} - 1} = \frac{π}{12} + \frac{π}{4}

\sqrt{e^{x} - 1} = t a n (\frac{π}{3})

e^{x} - 1 = 3

e^{x} = 4

x = l n 4 T a n m a y 16.07.19