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Question Number 116279 by mohammad17 last updated on 02/Oct/20

∫ ((xe^x )/((x+1)^2 ))dx

$$\int\:\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Oct/20

∫(((t−1)e^((t−1)) )/t^2 )dt=(1/e)∫e^t ((1/t)−(1/t^2 ))dt x+1=t =(1/e).e^t .(1/t)+C=(1/t)e^(t−1) +C=(e^x /(x+1))+C

$$\int\frac{\left(\mathrm{t}−\mathrm{1}\right)\mathrm{e}^{\left(\mathrm{t}−\mathrm{1}\right)} }{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{e}}\int\mathrm{e}^{\mathrm{t}} \left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}\:\:\:\:\:\:\:\mathrm{x}+\mathrm{1}=\mathrm{t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}.\mathrm{e}^{\mathrm{t}} .\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{C}=\frac{\mathrm{1}}{\mathrm{t}}\mathrm{e}^{\mathrm{t}−\mathrm{1}} +\mathrm{C}=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}+\mathrm{1}}+\mathrm{C} \\ $$

Commented by mohammad17 last updated on 02/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by TANMAY PANACEA last updated on 02/Oct/20

∫(((x+1)(d/dx)(e^x )−e^x (d/dx)(x+1))/((x+1)^2 ))dx ∫(d/dx)((e^x /(x+1)))dx=(e^x /(x+1))+c

$$\int\frac{\left({x}+\mathrm{1}\right)\frac{{d}}{{dx}}\left({e}^{{x}} \right)−{e}^{{x}} \frac{{d}}{{dx}}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{d}}{{dx}}\left(\frac{{e}^{{x}} }{{x}+\mathrm{1}}\right){dx}=\frac{{e}^{{x}} }{{x}+\mathrm{1}}+{c} \\ $$

Commented by mohammad17 last updated on 02/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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