∫(1/(e^(2x) −e^(−2x) )) dx


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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
$∫(1/(e^(2x) −(1/e^(2x) )))dx=∫(e^(2x) /(e^(2x) −1))dx= ∫(1+(1/(e^(2x) −1)))dx=x+∫(dx/(e^(2x) −1))=x+∫(dx/((e^x )^2 −1)) (1/((e^x )^2 −1))=(a/(e^x −1))+(b/(e^x +1))=(((a+b)e^x +(a−b))/((e^x )^2 −1)) ⇒ { ((a+b=0)),((a−b=1)) :}⇒ { ((a=(1/2))),((b=−(1/2))) :} (1/2)(∫(dx/(e^x −1))−∫(dx/(e^x +1)))=(1/2)(∫(dx/(e^x (1−(1/e^x ))))−∫(dx/(e^x (1+(1/e^x ))))) for firs integral set u=1−(1/e^x )⇒du=(1/e^x )dx for anothere integral set w=1+(1/e^x )⇒dw=−(1/e^x )dx so we have (1/2)(∫(du/u)+∫(dw/w))=(1/2)(lnu+lnw)=(1/2)(ln(1−(1/e^x ))+ln(1+(1/e^x )))= (1/2)ln(1−(1/e^(2x) ))$
$\int \frac{1}{e^{2 x} - \frac{1}{e^{2 x}}} d x = \int \frac{e^{2 x}}{e^{2 x} - 1} d x =$ $\int (1 + \frac{1}{e^{2 x} - 1}) d x = x + \int \frac{d x}{e^{2 x} - 1} = x + \int \frac{d x}{{(e^{x})}^{2} - 1}$ $\frac{1}{{(e^{x})}^{2} - 1} = \frac{a}{e^{x} - 1} + \frac{b}{e^{x} + 1} = \frac{(a + b) e^{x} + (a - b)}{{(e^{x})}^{2} - 1}$ $\Rightarrow {\begin{cases} a + b = 0 \\ a - b = 1 \end{cases} \Rightarrow {\begin{cases} a = \frac{1}{2} \\ b = - \frac{1}{2} \end{cases}$ $\frac{1}{2} (\int \frac{d x}{e^{x} - 1} - \int \frac{d x}{e^{x} + 1}) = \frac{1}{2} (\int \frac{d x}{e^{x} (1 - \frac{1}{e^{x}})} - \int \frac{d x}{e^{x} (1 + \frac{1}{e^{x}})})$ $f o r f i r s i n t e g r a l s e t u = 1 - \frac{1}{e^{x}} \Rightarrow d u = \frac{1}{e^{x}} d x$ $f o r a n o t h e r e i n t e g r a l s e t w = 1 + \frac{1}{e^{x}} \Rightarrow d w = - \frac{1}{e^{x}} d x$ $s o w e h a v e$ $\frac{1}{2} (\int \frac{d u}{u} + \int \frac{d w}{w}) = \frac{1}{2} (l n u + l n w) = \frac{1}{2} (l n (1 - \frac{1}{e^{x}}) + l n (1 + \frac{1}{e^{x}})) =$ $\frac{1}{2} l n (1 - \frac{1}{e^{2 x}})$
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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