Question Number 83927 by M±th+et£s last updated on 08/Mar/20
$$\int\frac{{sinh}\left({x}\right)+{e}^{\mathrm{3}{x}} }{{sinh}\left({x}\right)−{e}^{{x}} }\:{dx} \\ $$
Answered by TANMAY PANACEA last updated on 08/Mar/20
$$\int\frac{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}+{e}^{\mathrm{3}{x}} }{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}−{e}^{{x}} }{dx} \\ $$$$=\int\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{e}^{\mathrm{3}{x}} }{{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{e}^{{x}} }{dx} \\ $$$$=\int\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}+\mathrm{2}{e}^{\mathrm{4}{x}} }{−{e}^{\mathrm{2}{x}} −\mathrm{1}}{dx} \\ $$$$=\int\frac{\mathrm{1}−\mathrm{2}{e}^{\mathrm{4}{x}} −{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}}{dx} \\ $$$${t}={e}^{\mathrm{2}{x}} +\mathrm{1}\rightarrow{dt}={e}^{\mathrm{2}{x}} ×\mathrm{2}×{dx} \\ $$$$\frac{{dt}}{\mathrm{2}\left({t}−\mathrm{1}\right)}={dx} \\ $$$$\int\frac{\mathrm{1}−\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\left({t}−\mathrm{1}\right)}{{t}}×\frac{{dt}}{\mathrm{2}\left({t}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}−\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt}−\int\frac{{t}−\mathrm{1}}{{t}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}}−\int{dt}+\int\frac{{dt}}{{t}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}−\mathrm{1}\right)−{t}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({e}^{\mathrm{2}{x}} \right)−\left({e}^{\mathrm{2}{x}} +\mathrm{1}\right)+{c} \\ $$$${x}−\left({e}^{\mathrm{2}{x}} +\mathrm{1}\right)+{c} \\ $$
Commented by M±th+et£s last updated on 08/Mar/20
$${thank}\:{you}\:{sir}\:{nice}\:{solution} \\ $$