Question Number 186192 by normans last updated on 02/Feb/23
$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{My}}\:\boldsymbol{{old}}\:\boldsymbol{{problem}} \\ $$$$\:\:\:\int\:\:\boldsymbol{{e}}^{\boldsymbol{{tan}}\:\boldsymbol{{x}}} \:\:\boldsymbol{{dx}} \\ $$
Answered by MJS_new last updated on 02/Feb/23
![∫e^(tan x) dx= [t=tan x → dx=(dt/(t^2 +1))] =∫(e^t /(t^2 +1))dt=∫(e^t /((t−i)(t+i)))dt= =(i/2)∫((e^t /(t+i))−(e^t /(t−i)))dt= =(i/2)(e^(−i) Ei (t+i) −e^i Ei (t−1))= ... for the Integral Exponential Function Ei (x) do a Google search](https://www.tinkutara.com/question/Q186202.png)
$$\int\mathrm{e}^{\mathrm{tan}\:{x}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{\mathrm{e}^{{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\int\frac{\mathrm{e}^{{t}} }{\left({t}−\mathrm{i}\right)\left({t}+\mathrm{i}\right)}{dt}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\int\left(\frac{\mathrm{e}^{{t}} }{{t}+\mathrm{i}}−\frac{\mathrm{e}^{{t}} }{{t}−\mathrm{i}}\right){dt}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{e}^{−\mathrm{i}} \mathrm{Ei}\:\left({t}+\mathrm{i}\right)\:−\mathrm{e}^{\mathrm{i}} \mathrm{Ei}\:\left({t}−\mathrm{1}\right)\right)= \\ $$$$… \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{Integral}\:\mathrm{Exponential}\:\mathrm{Function}\:\mathrm{Ei}\:\left({x}\right) \\ $$$$\mathrm{do}\:\mathrm{a}\:\mathrm{Google}\:\mathrm{search} \\ $$