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Question Number 148567 by mathmax by abdo last updated on 29/Jul/21
find ∫_0 ^∞   ((x^2  logx)/((x^2 +1)^3 ))dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{logx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$
Answered by Ar Brandon last updated on 29/Jul/21
Ω=∫_0 ^∞ ((x^2 logx)/((x^2 +1)^3 ))dx      =∫_0 ^1 ((x^2 logx)/((x^2 +1)^3 ))dx+∫_1 ^∞ ((x^2 logx)/((x^2 +1)^3 ))dx      =∫_0 ^1 ((x^2 logx)/((x^2 +1)^3 ))dx−∫_0 ^1 ((x^2 logx)/((x^2 +1)^3 ))dx=0
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{log}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{log}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx}+\int_{\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{log}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{log}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{log}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx}=\mathrm{0} \\ $$

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