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Question-110549




Question Number 110549 by shahria14 last updated on 29/Aug/20
Commented by peter frank last updated on 29/Aug/20
Answered by peter frank last updated on 29/Aug/20
ans^  =(e^x /(1+x))
$$\mathrm{an}\overset{} {\mathrm{s}}\:=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{x}} \\ $$
Answered by Dwaipayan Shikari last updated on 29/Aug/20
∫((xe^x +e^x )/((1+x)^2 ))−(e^x /((1+x)^2 ))dx  ∫e^x ((1/(1+x))−(1/((1+x)^2 )))dx=(e^x /(1+x))+C
$$\int\frac{{xe}^{{x}} +{e}^{{x}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }−\frac{{e}^{{x}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$\int{e}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\right){dx}=\frac{{e}^{{x}} }{\mathrm{1}+{x}}+{C} \\ $$

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