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Question Number 37813 by prof Abdo imad last updated on 17/Jun/18
find A_n = ∫_(1/n) ^1 x(√x)arctan(x+(1/x))dx then calculate lim_(n→+∞) A_n .
$${find}\:{A}_{{n}} \:\:=\:\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\:{x}\sqrt{{x}}{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$${then}\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} . \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18
∫tan^(−1) (x+(1/x))×x^(3/2) dx tan^(−1) (x+(1/x))×(x^(5/2) /(5/2))−∫(1/(1+x^2 +2+(1/x^2 )))×(x^(5/2) /(5/2))dx do−(2/5)∫(x^(5/2) /(x^2 +(1/x^2 )+3))dx d0−(2/5)∫(x^(9/2) /(x^4 +3x^2 +1))dx x=t^2 dx=2tdt do−(2/5)∫((t^(2×(9/2)) ×2tdt)/(t^8 +3t^4 +1)) do−(4/5)∫(t^(10) /(t^8 +3t^4 +1))dt do−(4/5)∫((t^(10) +3t^6 +t^2 −3t^6 −t^2 )/(t^8 +3t^4 +1))dt do−(4/5)∫t^2 dt+(4/5)∫((3t^6 +t^2 )/(t^8 +3t^4 +1))dt do−(4/5)∫t^2 dt+(4/5)∫((3t^2 +(1/t^2 ))/(t^4 +3+(1/t^4 ))) do−(4/5)∫t^2 dt+(4/5)∫((3(t^2 +(1/t^2 ))−(2/t^2 ))/((t^2 +(1/t^2 ))^2 +1)) contd
$$\int{tan}^{−\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)×{x}^{\frac{\mathrm{3}}{\mathrm{2}}} {dx} \\ $$$${tan}^{−\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)×\frac{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\frac{\mathrm{5}}{\mathrm{2}}}−\int\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}×\frac{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\frac{\mathrm{5}}{\mathrm{2}}}{dx} \\ $$$${do}−\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{3}}{dx} \\ $$$${d}\mathrm{0}−\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{x}^{\frac{\mathrm{9}}{\mathrm{2}}} }{{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${x}={t}^{\mathrm{2}} \:\:\:{dx}=\mathrm{2}{tdt} \\ $$$${do}−\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{t}^{\mathrm{2}×\frac{\mathrm{9}}{\mathrm{2}}} ×\mathrm{2}{tdt}}{{t}^{\mathrm{8}} +\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}} \\ $$$${do}−\frac{\mathrm{4}}{\mathrm{5}}\int\frac{{t}^{\mathrm{10}} }{{t}^{\mathrm{8}} +\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$${do}−\frac{\mathrm{4}}{\mathrm{5}}\int\frac{{t}^{\mathrm{10}} +\mathrm{3}{t}^{\mathrm{6}} +{t}^{\mathrm{2}} −\mathrm{3}{t}^{\mathrm{6}} −{t}^{\mathrm{2}} }{{t}^{\mathrm{8}} +\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$${do}−\frac{\mathrm{4}}{\mathrm{5}}\int{t}^{\mathrm{2}} {dt}+\frac{\mathrm{4}}{\mathrm{5}}\int\frac{\mathrm{3}{t}^{\mathrm{6}} +{t}^{\mathrm{2}} }{{t}^{\mathrm{8}} +\mathrm{3}{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$${do}−\frac{\mathrm{4}}{\mathrm{5}}\int{t}^{\mathrm{2}} {dt}+\frac{\mathrm{4}}{\mathrm{5}}\int\frac{\mathrm{3}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{4}} +\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{4}} }} \\ $$$${do}−\frac{\mathrm{4}}{\mathrm{5}}\int{t}^{\mathrm{2}} {dt}+\frac{\mathrm{4}}{\mathrm{5}}\int\frac{\mathrm{3}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$${contd} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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