Question Number 14399 by tawa tawa last updated on 31/May/17
$$\mathrm{Evaluate}:\:\:\:\:\:\:\int\:\frac{\mathrm{1}\:+\:\mathrm{e}^{\mathrm{x}} \:−\:\mathrm{e}^{\mathrm{3x}} }{\mathrm{e}^{−\mathrm{x}} \:−\:\mathrm{e}^{\mathrm{x}} }\:\:\mathrm{dx} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17
$${e}^{{x}} ={t}\Rightarrow{tdx}={dt}\Rightarrow{dx}=\frac{{dt}}{{t}} \\ $$$${I}=\int\frac{\mathrm{1}+{t}−{t}^{\mathrm{3}} }{\frac{\mathrm{1}}{{t}}−{t}}.\frac{{dt}}{{t}}=\int\frac{{t}^{\mathrm{3}} −{t}−\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}= \\ $$$$\int\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}=\int{tdt}−\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{ln}\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}+{C}=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{x}} −{ln}\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}\right)+{C}\:.\blacksquare \\ $$
Commented by tawa tawa last updated on 31/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$