1-e-2x-e-2x-dx-

Question Number 61966 by aliesam last updated on 12/Jun/19

\int \frac{1}{e^{2 x} - e^{- 2 x}} d x

Commented by kaivan.ahmadi last updated on 12/Jun/19

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

\int (1 + \frac{1}{e^{2 x} - 1}) d x = x + \int \frac{d x}{e^{2 x} - 1} = x + \int \frac{d x}{{(e^{x})}^{2} - 1}

\frac{1}{{(e^{x})}^{2} - 1} = \frac{a}{e^{x} - 1} + \frac{b}{e^{x} + 1} = \frac{(a + b) e^{x} + (a - b)}{{(e^{x})}^{2} - 1}

\Rightarrow {\begin{cases} a + b = 0 \\ a - b = 1 \end{cases} \Rightarrow {\begin{cases} a = \frac{1}{2} \\ b = - \frac{1}{2} \end{cases}

\frac{1}{2} (\int \frac{d x}{e^{x} - 1} - \int \frac{d x}{e^{x} + 1}) = \frac{1}{2} (\int \frac{d x}{e^{x} (1 - \frac{1}{e^{x}})} - \int \frac{d x}{e^{x} (1 + \frac{1}{e^{x}})})

f o r f i r s i n t e g r a l s e t u = 1 - \frac{1}{e^{x}} \Rightarrow d u = \frac{1}{e^{x}} d x

f o r a n o t h e r e i n t e g r a l s e t w = 1 + \frac{1}{e^{x}} \Rightarrow d w = - \frac{1}{e^{x}} d x

s o w e h a v e

\frac{1}{2} (\int \frac{d u}{u} + \int \frac{d w}{w}) = \frac{1}{2} (l n u + l n w) = \frac{1}{2} (l n (1 - \frac{1}{e^{x}}) + l n (1 + \frac{1}{e^{x}})) =

\frac{1}{2} l n (1 - \frac{1}{e^{2 x}})

Commented by maxmathsup by imad last updated on 12/Jun/19

a t f o r m o f s e r i e

I = \int \frac{d x}{e^{2 x} - e^{- 2 x}} = \int \frac{e^{- 2 x}}{1 - e^{- 4 x}} d x = \int e^{- 2 x} (\sum_{n = 0}^{\infty} e^{- 4 n x} d x)

= \sum_{n = 0}^{\infty} \int e^{- (2 + 4 n) x} d x = \sum_{n = 0}^{\infty} A_{n} w i t h A_{n} = \int e^{- (4 n + 2) x} d x

= \frac{- 1}{4 n + 2} e^{- (4 n + 2)} \Rightarrow I = - \sum_{n = 0}^{\infty} \frac{1}{4 n + 2} e^{- (4 n + 2)} + C .

Commented by kaivan.ahmadi last updated on 12/Jun/19

I = \int \frac{1}{e^{2 x} - 1} d x = \int \frac{d x}{e^{2 x} (1 - \frac{1}{e^{2 x}})}

s e t u = 1 - \frac{1}{e^{2 x}} \Rightarrow d u = \frac{2 d x}{e^{2 x}}

I = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} l n u = \frac{1}{2} l n (1 - \frac{1}{e^{2 x}})

t h i s i s s h o r t w a y f o r t h i s i n t e g r a l

Commented by maxmathsup by imad last updated on 12/Jun/19

\frac{1}{e^{2 x} - \frac{1}{e^{2 x}}} = \frac{e^{2 x}}{e^{4 x} - 1} \neq \frac{e^{2 x}}{e^{2 x} - 1}

Answered by MJS last updated on 13/Jun/19

\int \frac{d x}{e^{2 x} - e^{- 2 x}} =

[t = e^{2 x} \to d x = \frac{d t}{2 t}]

= \int \frac{d t}{2 (t^{2} - 1)} = \int (\frac{1}{4 (t - 1)} - \frac{1}{4 (t + 1)}) d t = \frac{1}{4} \ln ∣ \frac{t - 1}{t + 1} ∣ =

= \frac{1}{4} \ln \frac{e^{2 x} - 1}{e^{2 x} + 1} + C = \frac{1}{4} \ln \tanh x + C

btw . \int \frac{d x}{e^{2 x} - e^{- 2 x}} = \int \frac{d x}{2 \sinh 2 x}

Answered by mr W last updated on 13/Jun/19

\int \frac{1}{e^{2 x} - e^{- 2 x}} d x

= \frac{1}{2} \int \frac{1}{e^{2 x} - e^{- 2 x}} d (2 x)

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

= \frac{1}{2} \int \frac{e^{t}}{{(e^{t})}^{2} - 1} d t

= \frac{1}{2} \int \frac{1}{{(e^{t})}^{2} - 1} d (e^{t})

= \frac{1}{2} \int \frac{1}{s^{2} - 1} d s

= \frac{1}{4} \int [\frac{1}{s - 1} - \frac{1}{s + 1}] d s

= \frac{1}{4} \ln ∣ \frac{s - 1}{s + 1} ∣ + C

= \frac{1}{4} \ln ∣ \frac{e^{2 x} - 1}{e^{2 x} + 1} ∣ + C

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