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Question Number 138355    Answers: 0   Comments: 1

Question Number 138350    Answers: 3   Comments: 0

(√2)((√8)−(2/( (√8))))

$$\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{8}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{8}}}\right) \\ $$

Question Number 138348    Answers: 0   Comments: 1

Question Number 138338    Answers: 1   Comments: 1

Question Number 138330    Answers: 2   Comments: 0

Question Number 138329    Answers: 3   Comments: 0

Question Number 138320    Answers: 1   Comments: 0

Question Number 138317    Answers: 1   Comments: 1

Question Number 138316    Answers: 0   Comments: 0

f(x)=(1/( (√(1+x))))+(1/( (√(1+a))))+(√((ax)/(ax+8))) ,x∈(0,∞) ∀ a∈(0,∞),show 1<f(x)<2.

$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}}}+\sqrt{\frac{{ax}}{{ax}+\mathrm{8}}}\:\:\:\:\:\:\:\:,{x}\in\left(\mathrm{0},\infty\right) \\ $$$$\forall\:{a}\in\left(\mathrm{0},\infty\right),{show}\:\mathrm{1}<{f}\left({x}\right)<\mathrm{2}. \\ $$

Question Number 138315    Answers: 3   Comments: 0

∫_(−∞) ^( +∞) ((cosx)/((x^2 +1)))dx don′t use feynmann trick

$$\int_{−\infty} ^{\:+\infty} \frac{\mathrm{cos}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{use}\:\mathrm{feynmann}\:\mathrm{trick} \\ $$

Question Number 138336    Answers: 1   Comments: 1

Question Number 138335    Answers: 1   Comments: 0

∫(dx/(x^8 +x^4 +1))=?

$$\int\frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=? \\ $$

Question Number 138384    Answers: 1   Comments: 0

find the singular point (1) (e^z /z^2 ) (2)((sin(z))/z) help me sir

$${find}\:{the}\:{singular}\:{point}\: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\frac{{e}^{{z}} }{{z}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\frac{{sin}\left({z}\right)}{{z}} \\ $$$$ \\ $$$${help}\:{me}\:{sir} \\ $$

Question Number 138302    Answers: 2   Comments: 0

if tan^2 x=1+2tan^2 y show that cos2x+sin^2 y=0

$${if}\:\:\mathrm{tan}^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{2tan}^{\mathrm{2}} {y} \\ $$$${show}\:{that} \\ $$$$\mathrm{cos2}{x}+\mathrm{sin}^{\mathrm{2}} {y}=\mathrm{0} \\ $$

Question Number 138299    Answers: 2   Comments: 0

Question Number 138296    Answers: 3   Comments: 0

∫ (dx/(x^4 (√(x^2 −a^2 )))) =?

$$\int\:\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=? \\ $$

Question Number 138295    Answers: 1   Comments: 0

hi ! Simplify : for n ∈ N^∗ , S_n = Σ_(k=1) ^n ((3k+8)/(k(k+2)2^k )) .

$$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{Simplify}}\::\:{for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\:{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\:\frac{\mathrm{3}{k}+\mathrm{8}}{{k}\left({k}+\mathrm{2}\right)\mathrm{2}^{{k}} }\:. \\ $$

Question Number 138283    Answers: 0   Comments: 2

......advanced ........... calculus...... prove that:: 𝛗=∫_0 ^( 1) ((ln(1+x^2 ).arctan(x))/x^2 )dx= proof::: 𝛗=_(⟨substitution⟩) ^(x=tan(θ)) ∫_0 ^( (π/4)) ((ln(1+tan^2 (θ)).θ)/(tan^2 (θ)))(1+tan^2 (θ))dθ =^(⟨simplification⟩) ∫_0 ^( (π/4)) ((θ.ln((1/(cos^2 (θ)))))/(sin^2 (θ)))dθ =−2∫_0 ^( (π/4)) ((θ.ln(cos(θ))/(sin^2 (θ)))dθ =^(i.b.p) 2{[(cot(θ).θ.ln(cos(θ))]_0 ^(π/4) −∫_0 ^( (π/4)) (cot(θ).[ln(cos(θ))−θ.tan(θ)]dθ =2.(π/4).ln(((√2)/2))−2∫_0 ^( (π/4)) cot(θ).ln(cos(θ))dθ+2∫_0 ^( (π/4)) θdθ =((−π)/4)ln(2)−Φ+(π^2 /(16)) Φ=∫_0 ^( (π/4)) ((cos(θ))/(sin(θ))).ln(1−sin^2 (θ))dθ =^(sin(θ)=y) ∫_0 ^( ((√2)/2)) ((ln(1−y^2 ))/y)dy=−∫_0 ^( ((√2)/2)) Σ_(n=1) ^∞ (y^(2n−1) /n)dy =−Σ[(y^(2n) /(2n^2 ))]_0 ^( ((√2)/2)) =((−1)/2) li_2 ((1/2)) =((−1)/2){(π^2 /(12))−(1/2)ln^2 (2)}=((−π^2 )/(24))+(1/4)ln^2 (2)... ∴ 𝛗=((−π)/4)ln(2)+(π^2 /(24))+(1/4)ln^2 (2)+(π^2 /(16)) ......... 𝛗 =((5π^2 )/(48))−((12π)/(48))ln(2)+((12)/(48)) ln^2 (2) ..... ........𝛗=(1/(48)){5π^2 −12πln(2)+12ln^2 (2)}

$$\:\:\:\:\:\:\:\:\:\:\:......{advanced}\:\:\:...........\:\:{calculus}...... \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right).{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}= \\ $$$$\:{proof}::: \\ $$$$\:\:\:\boldsymbol{\phi}\underset{\langle{substitution}\rangle} {\overset{{x}={tan}\left(\theta\right)} {=}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\theta\right)\right).\theta}{{tan}^{\mathrm{2}} \left(\theta\right)}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\theta\right)\right){d}\theta \\ $$$$\overset{\langle{simplification}\rangle} {=}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\theta.{ln}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\theta\right)}\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}{d}\theta \\ $$$$\:\:\:\:\:\:\:=−\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\theta.{ln}\left({cos}\left(\theta\right)\right.}{{sin}^{\mathrm{2}} \left(\theta\right)}{d}\theta \\ $$$$\:\:\:\:\:\overset{{i}.{b}.{p}} {=}\mathrm{2}\left\{\left[\left({cot}\left(\theta\right).\theta.{ln}\left({cos}\left(\theta\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left({cot}\left(\theta\right).\left[{ln}\left({cos}\left(\theta\right)\right)−\theta.{tan}\left(\theta\right)\right]{d}\theta\right.\right.\right. \\ $$$$\:\:\:\:\:\:=\mathrm{2}.\frac{\pi}{\mathrm{4}}.{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {cot}\left(\theta\right).{ln}\left({cos}\left(\theta\right)\right){d}\theta+\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \theta{d}\theta \\ $$$$\:\:\:\:\:\:=\frac{−\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\Phi+\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{cos}\left(\theta\right)}{{sin}\left(\theta\right)}.{ln}\left(\mathrm{1}−{sin}^{\mathrm{2}} \left(\theta\right)\right){d}\theta \\ $$$$\:\:\:\overset{{sin}\left(\theta\right)={y}} {=}\:\int_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{{y}}{dy}=−\int_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{y}^{\mathrm{2}{n}−\mathrm{1}} }{{n}}{dy} \\ $$$$\:\:\:\:\:\:=−\Sigma\left[\frac{{y}^{\mathrm{2}{n}} }{\mathrm{2}{n}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\frac{−\mathrm{1}}{\mathrm{2}}\:{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:=\frac{−\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right\}=\frac{−\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)... \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\boldsymbol{\phi}=\frac{−\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\:\:\:\:\:.........\:\:\:\:\boldsymbol{\phi}\:=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{12}\pi}{\mathrm{48}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{12}}{\mathrm{48}}\:{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:..... \\ $$$$\:\:\:\:\:\:\:\:\:\:........\boldsymbol{\phi}=\frac{\mathrm{1}}{\mathrm{48}}\left\{\mathrm{5}\pi^{\mathrm{2}} −\mathrm{12}\pi{ln}\left(\mathrm{2}\right)+\mathrm{12}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 138280    Answers: 0   Comments: 2

x^(x+4) =32 solution method?

$$\mathrm{x}^{\mathrm{x}+\mathrm{4}} =\mathrm{32}\:\:\:\:\mathrm{solution}\:\mathrm{method}? \\ $$

Question Number 138278    Answers: 1   Comments: 0

Question Number 138277    Answers: 3   Comments: 0

calculate ∫_0 ^∞ ∫_0 ^∞ e^(−x^2 −y^2 ) sin(x^2 +y^2 )dxdy

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$

Question Number 138275    Answers: 0   Comments: 0

1)calculate U_n =∫∫_([(1/n),1]^2 ) (2x+3y)(√(x^2 +y^2 ))dxdy 2)find ∫∫_(]0,1]^2 ) (2x+3y)(√(x^2 +y^2 ))dxdy

$$\left.\mathrm{1}\right)\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int\int_{\left[\frac{\mathrm{1}}{\mathrm{n}},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\int\int_{\left.\right]\left.\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$

Question Number 138276    Answers: 0   Comments: 0

1) calculate A_n =∫∫_([0,n[^2 ) ((dxdy)/((2x^2 +3y^2 )^2 )) 2)find lim_(n→+∞) A_n

$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{A}_{\mathrm{n}} =\int\int_{\left[\mathrm{0},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\frac{\mathrm{dxdy}}{\left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} \\ $$

Question Number 138240    Answers: 3   Comments: 0

(x−1)(dy/dx) +xy = 2xe^(−x)

$$\:\left({x}−\mathrm{1}\right)\frac{{dy}}{{dx}}\:+{xy}\:=\:\mathrm{2}{xe}^{−{x}} \\ $$

Question Number 138257    Answers: 1   Comments: 0

hi ! calculate : ∫∫_A (x^2 −y^2 )dxdy with A={(x^2 /a^2 ) + (y^2 /b^2 ) ≤ 1}

$$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\: \\ $$$$\int\int_{\mathrm{A}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:{with}\:\mathrm{A}=\left\{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\right\} \\ $$

Question Number 138254    Answers: 0   Comments: 10

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