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AllQuestion and Answers: Page 447

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Find A,B⊂N^∗ Such that A∩B=∅ , cardB=cardA+1≥8 and Σ_(x∈A) x^3 = Σ_(x∈B) x^3

$$\mathrm{Find}\:\:\mathrm{A},\mathrm{B}\subset\mathbb{N}^{\ast} \\ $$$$\mathrm{Such}\:\mathrm{that}\:\:\mathrm{A}\cap\mathrm{B}=\varnothing\:,\:\mathrm{cardB}=\mathrm{cardA}+\mathrm{1}\geqslant\mathrm{8} \\ $$$$\mathrm{and}\:\:\underset{\boldsymbol{\mathrm{x}}\in\boldsymbol{\mathrm{A}}} {\sum}\:\mathrm{x}^{\mathrm{3}} \:=\:\underset{\boldsymbol{\mathrm{x}}\in\boldsymbol{\mathrm{B}}} {\sum}\:\mathrm{x}^{\mathrm{3}} \\ $$

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let U_n =∫_0 ^1 (√(1−x^n ))ln^2 xdx 1)lim U_n ? 2)equivalent of U_n (n→∞)

$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{ln}^{\mathrm{2}} {xdx} \\ $$$$\left.\mathrm{1}\right){lim}\:{U}_{{n}} ? \\ $$$$\left.\mathrm{2}\right){equivalent}\:{of}\:{U}_{{n}} \left({n}\rightarrow\infty\right) \\ $$

Question Number 181403 Answers: 1 Comments: 0

Refer to Q181319 ∫((x^4 +1)/(x^4 +x+2))dx ((x^4 +1)/(x^4 +x+2))=1+(x/(2(x^2 +x+1)))−(x/(2(x^2 −x+1))) =1+(1/4)[(((2x+1)−1)/(x^2 +x+1))−(1/4)×(((2x−1)+1)/((x^2 −x+1))) =(1/4)×[log(x^2 +x+1)]^′ −(1/4)×((1/([((√3)/2)(x+(1/2))^2 ]+1))) −((1/4)[log(x^2 −x+1)]^′ +(1/4)((1/([((√3)/2)×(x−(1/2)]^2 +1)))) first with [((√3)/2)( x+(1/2))]=u ∫(1/(u^2 +1))du=arctg(u) and v=[((√3)/2)(x−(1/2) )]=v ∫(dv/(1+v^2 ))=arctg(v) ...............

$${Refer}\:{to}\:\mathrm{Q181319} \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}+\mathrm{2}}{dx} \\ $$$$\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}+\mathrm{2}}=\mathrm{1}+\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left[\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\right]^{'} −\frac{\mathrm{1}}{\mathrm{4}}×\left(\frac{\mathrm{1}}{\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right]+\mathrm{1}}\right) \\ $$$$−\left(\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)\right]^{'} +\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right]^{\mathrm{2}} +\mathrm{1}\right.}\right)\right) \\ $$$${first}\:{with}\:\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right]={u}\:\:\:\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}={arctg}\left({u}\right) \\ $$$${and}\:{v}=\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\:\right)\right]=\mathrm{v}\:\:\:\int\frac{{dv}}{\mathrm{1}+{v}^{\mathrm{2}} }={arctg}\left({v}\right) \\ $$$$............... \\ $$

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calcul: R=+oo Σ_(n=o) ^(+oo) ((2n)/((2n+1)!))x^n

$${calcul}:\:{R}=+{oo}\: \\ $$$$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\frac{\mathrm{2}{n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{{n}} \\ $$

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∫_0 ^( ∞) (( 1)/((1+ x^( 4) ) (1+ x^( 6) )))dx=((p(√2) −q)/(12)) π p , q=?

$$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{1}}{\left(\mathrm{1}+\:{x}^{\:\mathrm{4}} \right)\:\left(\mathrm{1}+\:{x}^{\:\mathrm{6}} \right)}{dx}=\frac{{p}\sqrt{\mathrm{2}}\:−{q}}{\mathrm{12}}\:\pi \\ $$$$\:\:\:{p}\:,\:\:{q}=? \\ $$$$ \\ $$

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