Question Number 120919 by bounhome last updated on 04/Nov/20 | ||
$$\int\frac{\mathrm{1}}{\left({e}^{{x}} +\mathrm{1}\right)}{dx}=? \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 04/Nov/20 | ||
$$\int\frac{\mathrm{1}}{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=\int\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}\:\:\:\:\:\:\:\:\:\:\:\:{t}={e}^{{x}} +\mathrm{1}\Rightarrow\mathrm{1}={e}^{{x}} \frac{{dt}}{{dx}} \\ $$$$={log}\left(\frac{{t}−\mathrm{1}}{{t}}\right)+{C} \\ $$$$={log}\left(\frac{{e}^{{x}} }{{e}^{{x}} +\mathrm{1}}\right)+\mathrm{C} \\ $$ | ||
Commented by bounhome last updated on 04/Nov/20 | ||
$${thank}\:{you} \\ $$ | ||
Answered by liberty last updated on 04/Nov/20 | ||
$$\:\int\:\frac{\mathrm{e}^{\mathrm{x}} \:\mathrm{dx}}{\mathrm{e}^{\mathrm{x}} \left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)}\:=\:\int\:\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}} \right)}{\mathrm{e}^{\mathrm{x}} \left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)}\:=\:\int\:\frac{\mathrm{dr}}{\mathrm{r}\left(\mathrm{r}+\mathrm{1}\right)} \\ $$$$=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{r}}−\frac{\mathrm{1}}{\mathrm{r}+\mathrm{1}}\right)\mathrm{dr}\:=\:\mathrm{ln}\:\left(\frac{\mathrm{r}}{\mathrm{r}+\mathrm{1}}\right)+\mathrm{C} \\ $$$$=\:\mathrm{ln}\:\left(\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}\right)+\mathrm{c}\:.\blacktriangle \\ $$ | ||
Commented by bounhome last updated on 04/Nov/20 | ||
$${thank}\:{you} \\ $$ | ||