Question Number 116279 by mohammad17 last updated on 02/Oct/20
$$\int\:\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Oct/20
$$\int\frac{\left(\mathrm{t}−\mathrm{1}\right)\mathrm{e}^{\left(\mathrm{t}−\mathrm{1}\right)} }{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{e}}\int\mathrm{e}^{\mathrm{t}} \left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}\:\:\:\:\:\:\:\mathrm{x}+\mathrm{1}=\mathrm{t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}.\mathrm{e}^{\mathrm{t}} .\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{C}=\frac{\mathrm{1}}{\mathrm{t}}\mathrm{e}^{\mathrm{t}−\mathrm{1}} +\mathrm{C}=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}+\mathrm{1}}+\mathrm{C} \\ $$
Commented by mohammad17 last updated on 02/Oct/20
$${thank}\:{you}\:{sir} \\ $$
Answered by TANMAY PANACEA last updated on 02/Oct/20
$$\int\frac{\left({x}+\mathrm{1}\right)\frac{{d}}{{dx}}\left({e}^{{x}} \right)−{e}^{{x}} \frac{{d}}{{dx}}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{d}}{{dx}}\left(\frac{{e}^{{x}} }{{x}+\mathrm{1}}\right){dx}=\frac{{e}^{{x}} }{{x}+\mathrm{1}}+{c} \\ $$
Commented by mohammad17 last updated on 02/Oct/20
$${thank}\:{you}\:{sir} \\ $$