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Question Number 83927 by M±th+et£s last updated on 08/Mar/20

∫((sinh(x)+e^(3x) )/(sinh(x)−e^x )) dx

$$\int\frac{{sinh}\left({x}\right)+{e}^{\mathrm{3}{x}} }{{sinh}\left({x}\right)−{e}^{{x}} }\:{dx} \\ $$

Answered by TANMAY PANACEA last updated on 08/Mar/20

∫((((e^x −e^(−x) )/2)+e^(3x) )/(((e^x −e^(−x) )/2)−e^x ))dx =∫((e^x −e^(−x) +2e^(3x) )/(e^x −e^(−x) −2e^x ))dx =∫((e^(2x) −1+2e^(4x) )/(−e^(2x) −1))dx =∫((1−2e^(4x) −e^(2x) )/(e^(2x) +1))dx t=e^(2x) +1→dt=e^(2x) ×2×dx (dt/(2(t−1)))=dx ∫((1−2(t−1)^2 −(t−1))/t)×(dt/(2(t−1))) (1/2)∫((1−2(t−1)^2 −(t−1))/(t(t−1)))dt (1/2)∫((t−(t−1))/(t(t−1)))dt−∫((t−1)/t)dt−(1/2)∫(dt/t) (1/2)∫(dt/(t−1))−(1/2)∫(dt/t)−∫dt+∫(dt/t)−(1/2)∫(dt/t) (1/2)ln(t−1)−t+c (1/2)ln(e^(2x) )−(e^(2x) +1)+c x−(e^(2x) +1)+c

$$\int\frac{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}+{e}^{\mathrm{3}{x}} }{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}−{e}^{{x}} }{dx} \\ $$$$=\int\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{e}^{\mathrm{3}{x}} }{{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{e}^{{x}} }{dx} \\ $$$$=\int\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}+\mathrm{2}{e}^{\mathrm{4}{x}} }{−{e}^{\mathrm{2}{x}} −\mathrm{1}}{dx} \\ $$$$=\int\frac{\mathrm{1}−\mathrm{2}{e}^{\mathrm{4}{x}} −{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}}{dx} \\ $$$${t}={e}^{\mathrm{2}{x}} +\mathrm{1}\rightarrow{dt}={e}^{\mathrm{2}{x}} ×\mathrm{2}×{dx} \\ $$$$\frac{{dt}}{\mathrm{2}\left({t}−\mathrm{1}\right)}={dx} \\ $$$$\int\frac{\mathrm{1}−\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\left({t}−\mathrm{1}\right)}{{t}}×\frac{{dt}}{\mathrm{2}\left({t}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}−\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt}−\int\frac{{t}−\mathrm{1}}{{t}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}}−\int{dt}+\int\frac{{dt}}{{t}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}−\mathrm{1}\right)−{t}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({e}^{\mathrm{2}{x}} \right)−\left({e}^{\mathrm{2}{x}} +\mathrm{1}\right)+{c} \\ $$$${x}−\left({e}^{\mathrm{2}{x}} +\mathrm{1}\right)+{c} \\ $$

Commented by M±th+et£s last updated on 08/Mar/20

thank you sir nice solution

$${thank}\:{you}\:{sir}\:{nice}\:{solution} \\ $$

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