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

∫(1/(e^(2x) −e^(−2x) )) dx

$$\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }\:{dx} \\ $$

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) ))

$$\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }}{dx}=\int\frac{{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{2}{x}} −\mathrm{1}}{dx}= \\ $$$$\int\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −\mathrm{1}}\right){dx}={x}+\int\frac{{dx}}{{e}^{\mathrm{2}{x}} −\mathrm{1}}={x}+\int\frac{{dx}}{\left({e}^{{x}} \right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\left({e}^{{x}} \right)^{\mathrm{2}} −\mathrm{1}}=\frac{{a}}{{e}^{{x}} −\mathrm{1}}+\frac{{b}}{{e}^{{x}} +\mathrm{1}}=\frac{\left({a}+{b}\right){e}^{{x}} +\left({a}−{b}\right)}{\left({e}^{{x}} \right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\begin{cases}{{a}+{b}=\mathrm{0}}\\{{a}−{b}=\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{{a}=\frac{\mathrm{1}}{\mathrm{2}}}\\{{b}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{dx}}{{e}^{{x}} −\mathrm{1}}−\int\frac{{dx}}{{e}^{{x}} +\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{dx}}{{e}^{{x}} \left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{{x}} }\right)}−\int\frac{{dx}}{{e}^{{x}} \left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}} }\right)}\right) \\ $$$${for}\:{firs}\:{integral}\:{set}\:\:{u}=\mathrm{1}−\frac{\mathrm{1}}{{e}^{{x}} }\Rightarrow{du}=\frac{\mathrm{1}}{{e}^{{x}} }{dx} \\ $$$${for}\:\:{anothere}\:{integral}\:{set}\:{w}=\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}} }\Rightarrow{dw}=−\frac{\mathrm{1}}{{e}^{{x}} }{dx} \\ $$$${so}\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{du}}{{u}}+\int\frac{{dw}}{{w}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({lnu}+{lnw}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{{x}} }\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}} }\right)\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }\right) \\ $$$$ \\ $$

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

at form of serie I =∫ (dx/(e^(2x) −e^(−2x) )) =∫ (e^(−2x) /(1−e^(−4x) )) dx = ∫ e^(−2x) (Σ_(n=0) ^∞ e^(−4nx) dx) =Σ_(n=0) ^∞ ∫ e^(−(2+4n)x) dx =Σ_(n=0) ^∞ A_n with A_n =∫ e^(−(4n+2)x) dx =((−1)/(4n+2)) e^(−(4n+2)) ⇒ I =−Σ_(n=0) ^∞ (1/(4n+2)) e^(−(4n+2)) +C .

$${at}\:{form}\:{of}\:{serie} \\ $$$${I}\:=\int\:\:\:\:\frac{{dx}}{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }\:=\int\:\:\frac{{e}^{−\mathrm{2}{x}} }{\mathrm{1}−{e}^{−\mathrm{4}{x}} }\:{dx}\:=\:\int\:{e}^{−\mathrm{2}{x}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{4}{nx}} {dx}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int\:{e}^{−\left(\mathrm{2}+\mathrm{4}{n}\right){x}} {dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:\:\:{with}\:{A}_{{n}} =\int\:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}\right){x}} {dx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}{n}+\mathrm{2}}\:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}\right)} \:\Rightarrow\:{I}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{2}}\:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}\right)} \:\:+{C}\:. \\ $$

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

I=∫(1/(e^(2x) −1))dx=∫(dx/(e^(2x) (1−(1/e^(2x) )))) set u=1−(1/e^(2x) )⇒du=((2dx)/e^(2x) ) I=(1/2)∫(du/u)=(1/2)lnu=(1/2)ln(1−(1/e^(2x) )) this is short way for this integral

$${I}=\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −\mathrm{1}}{dx}=\int\frac{{dx}}{{e}^{\mathrm{2}{x}} \left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }\right)} \\ $$$${set}\:\:{u}=\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }\Rightarrow{du}=\frac{\mathrm{2}{dx}}{{e}^{\mathrm{2}{x}} } \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}}=\frac{\mathrm{1}}{\mathrm{2}}{lnu}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }\right) \\ $$$${this}\:{is}\:{short}\:{way}\:{for}\:{this}\:{integral} \\ $$

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

(1/(e^(2x) −(1/e^(2x) ))) =(e^(2x) /(e^(4x) −1)) ≠ (e^(2x) /(e^(2x) −1))

$$\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} }}\:=\frac{{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{4}{x}} −\mathrm{1}}\:\neq\:\frac{{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{2}{x}} −\mathrm{1}} \\ $$

Answered by MJS last updated on 13/Jun/19

∫(dx/(e^(2x) −e^(−2x) ))= [t=e^(2x) → dx=(dt/(2t))] =∫(dt/(2(t^2 −1)))=∫((1/(4(t−1)))−(1/(4(t+1))))dt=(1/4)ln ∣((t−1)/(t+1))∣ = =(1/4)ln ((e^(2x) −1)/(e^(2x) +1)) +C=(1/4)ln tanh x +C btw. ∫(dx/(e^(2x) −e^(−2x) ))=∫(dx/(2sinh 2x))

$$\int\frac{{dx}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{e}^{−\mathrm{2}{x}} }= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{e}^{\mathrm{2}{x}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{t}}\right] \\ $$$$=\int\frac{{dt}}{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}=\int\left(\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({t}+\mathrm{1}\right)}\right){dt}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}\:+{C}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{tanh}\:{x}\:+{C} \\ $$$$\mathrm{btw}.\:\int\frac{{dx}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{e}^{−\mathrm{2}{x}} }=\int\frac{{dx}}{\mathrm{2sinh}\:\mathrm{2}{x}} \\ $$

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

∫(1/(e^(2x) −e^(−2x) )) dx =(1/2)∫(1/(e^(2x) −e^(−2x) )) d(2x) =(1/2)∫(1/(e^t −e^(−t) )) dt =(1/2)∫(e^t /((e^t )^2 −1)) dt =(1/2)∫(1/((e^t )^2 −1)) d(e^t ) =(1/2)∫(1/(s^2 −1)) ds =(1/4)∫[(1/(s−1))−(1/(s+1))] ds =(1/4)ln ∣((s−1)/(s+1))∣+C =(1/4)ln ∣((e^(2x) −1)/(e^(2x) +1))∣+C

$$\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }\:{d}\left(\mathrm{2}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{e}^{{t}} −{e}^{−{t}} }\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{e}^{{t}} }{\left({e}^{{t}} \right)^{\mathrm{2}} −\mathrm{1}}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({e}^{{t}} \right)^{\mathrm{2}} −\mathrm{1}}\:{d}\left({e}^{{t}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{s}^{\mathrm{2}} −\mathrm{1}}\:{ds} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left[\frac{\mathrm{1}}{{s}−\mathrm{1}}−\frac{\mathrm{1}}{{s}+\mathrm{1}}\right]\:{ds} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{{s}−\mathrm{1}}{{s}+\mathrm{1}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}\mid+{C} \\ $$

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