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Evaluate-e-x-1-x-1-x-2-2-dx-




Question Number 144896 by physicstutes last updated on 30/Jun/21
Evaluate    ∫e^x (((1−x)/(1+x^2 )))^2 dx
$$\mathrm{Evaluate}\: \\ $$$$\:\int{e}^{{x}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} {dx}\: \\ $$
Answered by Dwaipayan Shikari last updated on 30/Jun/21
(d/dx)(f(x)e^x )=f(x)e^x +f′(x)e^x   f(x)e^x =∫f(x)e^x +f′(x)e^x dx  ∫e^x ((1/(1+x^2 ))−((2x)/((1+x^2 )^2 )))dx=(e^x /(1+x^2 ))+C
$$\frac{{d}}{{dx}}\left({f}\left({x}\right){e}^{{x}} \right)={f}\left({x}\right){e}^{{x}} +{f}'\left({x}\right){e}^{{x}} \\ $$$${f}\left({x}\right){e}^{{x}} =\int{f}\left({x}\right){e}^{{x}} +{f}'\left({x}\right){e}^{{x}} {dx} \\ $$$$\int{e}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){dx}=\frac{{e}^{{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }+{C} \\ $$

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