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Question Number 73180    Answers: 0   Comments: 0

calculate ∫_0 ^∞ ((lnx)/((x+1)^3 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 73179    Answers: 1   Comments: 1

caoculate ∫_0 ^∞ ((arctan(x^2 −1))/(2x^2 +1))dx

$${caoculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$

Question Number 73178    Answers: 1   Comments: 0

calculate ∫_0 ^∞ ((ln(2+x^2 ))/(x^2 −x+1))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$

Question Number 73160    Answers: 1   Comments: 0

1+tanAtan(A/2)=tanAcot(A/2)−1=secA

$$\mathrm{1}+{tanAtan}\frac{{A}}{\mathrm{2}}={tanAcot}\frac{{A}}{\mathrm{2}}−\mathrm{1}={secA} \\ $$

Question Number 73155    Answers: 0   Comments: 0

reposting a former question... ∫(((x)^(1/5) −1)/((√x)+1))dx= [t=(x)^(1/(10)) → dx=10(x^9 )^(1/(10)) dx] =10∫((t^9 (t−1))/(t^4 −t^3 +t^2 −t+1))dt= =10∫(t^6 −t^4 −t)dt+10∫((t(t^2 −t+1))/(t^4 −t^3 +t^2 −t+1))dt= =((10)/7)t^7 −2t^5 −5t^2 +(5+(√5))∫(t/(t^2 −((1−(√5))/3)t+1))dt+(5−(√5))∫(t/(t^2 −((1+(√5))/2)t+1))dt= and it′s easy to solve these

$$\mathrm{reposting}\:\mathrm{a}\:\mathrm{former}\:\mathrm{question}... \\ $$$$\int\frac{\sqrt[{\mathrm{5}}]{{x}}−\mathrm{1}}{\sqrt{{x}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{10}}]{{x}}\:\rightarrow\:{dx}=\mathrm{10}\sqrt[{\mathrm{10}}]{{x}^{\mathrm{9}} }{dx}\right] \\ $$$$=\mathrm{10}\int\frac{{t}^{\mathrm{9}} \left({t}−\mathrm{1}\right)}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\mathrm{10}\int\left({t}^{\mathrm{6}} −{t}^{\mathrm{4}} −{t}\right){dt}+\mathrm{10}\int\frac{{t}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{10}}{\mathrm{7}}{t}^{\mathrm{7}} −\mathrm{2}{t}^{\mathrm{5}} −\mathrm{5}{t}^{\mathrm{2}} +\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)\int\frac{{t}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{3}}{t}+\mathrm{1}}{dt}+\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)\int\frac{{t}}{{t}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{t}+\mathrm{1}}{dt}= \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\ $$

Question Number 73147    Answers: 1   Comments: 1

∫((x−6)/(x^3 +1))dx

$$\int\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$$$ \\ $$

Question Number 73144    Answers: 1   Comments: 1

calculte ∫ ((x+(√(2+x^2 )))/(x+1−(√(2+x^2 ))))dx

$${calculte}\:\int\:\:\frac{{x}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{dx} \\ $$

Question Number 73137    Answers: 0   Comments: 8

Question Number 73131    Answers: 0   Comments: 2

solve for x,in terms of: a∈R . x+(√x)+(√(x^2 −a))+(√(x−a^2 ))=a^2

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\:. \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$

Question Number 73117    Answers: 2   Comments: 1

Question Number 73111    Answers: 1   Comments: 0

Question Number 73113    Answers: 1   Comments: 3

Question Number 73090    Answers: 2   Comments: 1

Question Number 73211    Answers: 0   Comments: 0

y=(c+3)(√x) +((3+d)/x)−((a+4)/x^(a+3) )

$${y}=\left({c}+\mathrm{3}\right)\sqrt{{x}}\:+\frac{\mathrm{3}+{d}}{{x}}−\frac{{a}+\mathrm{4}}{{x}^{{a}+\mathrm{3}} } \\ $$

Question Number 73080    Answers: 1   Comments: 0

y′′=e^y pls solve

$$\mathrm{y}''=\mathrm{e}^{\mathrm{y}} \\ $$$$\mathrm{pls}\:\mathrm{solve} \\ $$

Question Number 73059    Answers: 1   Comments: 3

let P_n (x)=(x+1)^n −(x−1)^n 1) fartorize inside C(x) P_n (x) 2)calculate Π_(k=1) ^p cotan(((kπ)/(2p+1)))

$${let}\:{P}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} −\left({x}−\mathrm{1}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{fartorize}\:{inside}\:{C}\left({x}\right)\:{P}_{{n}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\prod_{{k}=\mathrm{1}} ^{{p}} \:{cotan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right) \\ $$

Question Number 73057    Answers: 1   Comments: 0

let P_n =(x+1)^(2n+1) −x^(2n+1) −1 prove that x^2 +x divide P_n

$${let}\:{P}_{{n}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} −{x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1} \\ $$$${prove}\:{that}\:{x}^{\mathrm{2}} \:+{x}\:{divide}\:\underset{{n}} {{P}} \\ $$

Question Number 73056    Answers: 0   Comments: 1

if (xsina+cosa)^n =q(x^2 +1)+r find r

$${if}\:\left({xsina}+{cosa}\right)^{{n}} ={q}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)+{r}\:{find}\:{r} \\ $$

Question Number 73055    Answers: 0   Comments: 0

decompose inside R(x) the fraction ((x^4 +x+1)/(x(x^2 +1)^3 ))

$${decompose}\:{inside}\:{R}\left({x}\right)\:{the}\:{fraction}\:\:\frac{{x}^{\mathrm{4}} \:+{x}+\mathrm{1}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 73052    Answers: 1   Comments: 0

let P_n =X^n +X^(n−1) +....+X^2 +X−1 ∈R[X] 1)prove that P_n have one root x_n inside ]0,+∞[ 2)study the sequence x_n

$${let}\:{P}_{{n}} ={X}^{{n}} \:+{X}^{{n}−\mathrm{1}} \:+....+{X}^{\mathrm{2}} \:+{X}−\mathrm{1}\:\in{R}\left[{X}\right] \\ $$$$\left.\mathrm{1}\left.\right){prove}\:{that}\:{P}_{{n}} {have}\:{one}\:{root}\:{x}_{{n}} \:{inside}\:\right]\mathrm{0},+\infty\left[\right. \\ $$$$\left.\mathrm{2}\right){study}\:{the}\:{sequence}\:{x}_{{n}} \\ $$

Question Number 73051    Answers: 0   Comments: 1

factorize inside R[X] 1)X^5 −1 2)X^6 +1

$${factorize}\:{inside}\:{R}\left[{X}\right] \\ $$$$\left.\mathrm{1}\right){X}^{\mathrm{5}} −\mathrm{1}\:\: \\ $$$$\left.\mathrm{2}\right){X}^{\mathrm{6}} \:+\mathrm{1} \\ $$

Question Number 73049    Answers: 0   Comments: 0

simplifyA_n = Σ_(k=0) ^n ((k/n)−α)^2 C_n ^k X^k (1−X)^(n−k)

$${simplifyA}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\frac{{k}}{{n}}−\alpha\right)^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{X}^{{k}} \left(\mathrm{1}−{X}\right)^{{n}−{k}} \\ $$

Question Number 73048    Answers: 0   Comments: 0

solve inside Z^2 x^2 +3xy−2y^2 =122

$${solve}\:{inside}\:{Z}^{\mathrm{2}} \:\:{x}^{\mathrm{2}} \:+\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\mathrm{122} \\ $$

Question Number 73047    Answers: 1   Comments: 0

solve inside Z^3 x^2 +y^2 +z^(2 ) =2xyz

$${solve}\:{inside}\:{Z}^{\mathrm{3}} \:\:\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}\:} =\mathrm{2}{xyz} \\ $$

Question Number 73046    Answers: 1   Comments: 0

prove that ∀n ∈N Σ_(k=0) ^n k C_(2n) ^(n+k) =nC_(2n−1) ^n

$${prove}\:{that}\:\forall{n}\:\in{N}\:\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\:{C}_{\mathrm{2}{n}} ^{{n}+{k}} \:={nC}_{\mathrm{2}{n}−\mathrm{1}} ^{{n}} \\ $$

Question Number 73045    Answers: 0   Comments: 0

calculate Σ_(k=0) ^n (C_n ^k /C_(2n−1) ^k )

$${calculate}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{C}_{{n}} ^{{k}} }{{C}_{\mathrm{2}{n}−\mathrm{1}} ^{{k}} } \\ $$

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