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Question Number 138579    Answers: 1   Comments: 0

Question Number 138576    Answers: 1   Comments: 0

lim_(x→0) (1+cot 2x)^(2tan 2x) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{cot}\:\mathrm{2}{x}\right)^{\mathrm{2tan}\:\mathrm{2}{x}} \:=?\: \\ $$

Question Number 138575    Answers: 0   Comments: 5

find x x^2 −2^x =1 Any help

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}^{\boldsymbol{\mathrm{x}}} =\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}} \\ $$

Question Number 138571    Answers: 1   Comments: 0

{ ((u_1 =1)),((u_(n+1) = ((n^2 −n+1)/n^2 ))) :} ; ∀n∈R lim_(n→∞) u_n =?

$$\begin{cases}{{u}_{\mathrm{1}} =\mathrm{1}}\\{{u}_{{n}+\mathrm{1}} =\:\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}\end{cases}\:;\:\forall{n}\in\mathbb{R} \\ $$$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=? \\ $$

Question Number 138570    Answers: 1   Comments: 0

∫_( (√2)) ^2 (dx/(x^2 (√((x^2 −1)^3 )))) =?

$$\underset{\:\sqrt{\mathrm{2}}} {\overset{\mathrm{2}} {\int}}\:\frac{{dx}}{{x}^{\mathrm{2}} \:\sqrt{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }}\:=?\: \\ $$

Question Number 138569    Answers: 1   Comments: 0

Question Number 138564    Answers: 1   Comments: 0

y∙y′=0.5∙(1+y∙c_1 )^2 ∙c_2 +0.5 y=?

$${y}\centerdot{y}'=\mathrm{0}.\mathrm{5}\centerdot\left(\mathrm{1}+{y}\centerdot{c}_{\mathrm{1}} \right)^{\mathrm{2}} \centerdot{c}_{\mathrm{2}} +\mathrm{0}.\mathrm{5} \\ $$$$ \\ $$$${y}=? \\ $$

Question Number 138560    Answers: 1   Comments: 0

Σ_(n=1) ^∞ (−1)^n (ln (n+1)−ln (n))=?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{ln}\:\left({n}+\mathrm{1}\right)−\mathrm{ln}\:\left({n}\right)\right)=? \\ $$

Question Number 138554    Answers: 1   Comments: 0

Question Number 138552    Answers: 0   Comments: 0

...nice mathemayics ... 𝛗=∫_0 ^( ∞) ((sin(tan(x)))/x)dx=(π/2)(1−(1/e)) .......

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:\:\:\:\:\:{mathemayics}\:... \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({tan}\left({x}\right)\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$$$\:\:....... \\ $$

Question Number 138548    Answers: 1   Comments: 0

Question Number 138546    Answers: 1   Comments: 0

Question Number 138539    Answers: 0   Comments: 0

find the intigral of complex number I_j =∫_(γj) xdz if j=1,2 and Y_1 he is a circle ∣z∣=R

$${find}\:{the}\:{intigral}\:{of}\:{complex}\:{number} \\ $$$$ \\ $$$${I}_{{j}} =\int_{\gamma{j}} {xdz}\:\:\:{if}\:{j}=\mathrm{1},\mathrm{2}\:{and}\:{Y}_{\mathrm{1}} {he}\:{is}\:{a}\:{circle} \\ $$$$ \\ $$$$\mid{z}\mid={R} \\ $$

Question Number 138536    Answers: 0   Comments: 1

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$$\mathrm{This}\:\mathrm{app}\:\mathrm{is}\:\mathrm{now}\:\mathrm{free}\:\mathrm{as}\:\mathrm{several}\:\mathrm{users} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{have}\:\mathrm{cards}\:\mathrm{and}\:\mathrm{are}\:\mathrm{not}\:\mathrm{able} \\ $$$$\mathrm{to}\:\mathrm{upgrade}.\:\mathrm{Manual}\:\mathrm{entry}\:\mathrm{in}\:\mathrm{backend} \\ $$$$\mathrm{system}\:\mathrm{had}\:\mathrm{put}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{overhead}. \\ $$$$\mathrm{Hence}\:\mathrm{we}\:\mathrm{have}\:\mathrm{decide}\:\mathrm{to}\:\mathrm{make}\:\mathrm{this} \\ $$$$\mathrm{app}\:\mathrm{free}. \\ $$$$\mathrm{We}\:\mathrm{issued}\:\mathrm{several}\:\mathrm{refunds}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{who}\:\mathrm{had}\:\mathrm{bought}.\:\mathrm{However}\:\mathrm{Google} \\ $$$$\mathrm{playstore}\:\mathrm{is}\:\mathrm{not}\:\mathrm{letting}\:\mathrm{us}\:\mathrm{issue} \\ $$$$\mathrm{more}\:\mathrm{refunds}\:\mathrm{as}\:\mathrm{some}\:\mathrm{security}\:\mathrm{check} \\ $$$$\mathrm{has}\:\mathrm{detected}\:\mathrm{too}\:\mathrm{many}\:\mathrm{refunds}\:\mathrm{issued} \\ $$$$\mathrm{and}\:\mathrm{blocked}\:\mathrm{further}\:\mathrm{refunds}. \\ $$$$\mathrm{Please}\:\mathrm{send}\:\mathrm{us}\:\mathrm{email}\:\mathrm{with}\:\mathrm{your} \\ $$$$\mathrm{order}\:\mathrm{id}\:\mathrm{and}\:\mathrm{we}\:\mathrm{will}\:\mathrm{those}\:\mathrm{refunds} \\ $$$$\mathrm{on}\:\mathrm{priority}\:\mathrm{as}\:\mathrm{soon}\:\mathrm{as}\:\mathrm{Google}\:\mathrm{playstore} \\ $$$$\mathrm{allows}\:\mathrm{us}. \\ $$

Question Number 138533    Answers: 0   Comments: 2

∫_0 ^5 (1+x)δ(x^2 −4)dx=?

$$\int_{\mathrm{0}} ^{\mathrm{5}} \left(\mathrm{1}+{x}\right)\delta\left({x}^{\mathrm{2}} −\mathrm{4}\right){dx}=? \\ $$

Question Number 138532    Answers: 2   Comments: 0

x+(√y)=7 (√x)+y=11 faind x=? and y=?

$${x}+\sqrt{{y}}=\mathrm{7} \\ $$$$\sqrt{{x}}+{y}=\mathrm{11} \\ $$$${faind}\:\:{x}=?\:\:{and}\:\:{y}=? \\ $$

Question Number 138524    Answers: 3   Comments: 0

.......advanced ... .... ... calculus..... I:=∫_((−π)/2) ^( (π/2)) sin^2 (tan(x))dx=^(???) (π/e)sinh(1)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:.......{advanced}\:...\:....\:...\:{calculus}..... \\ $$$$\:\:\:\boldsymbol{\mathrm{I}}:=\int_{\frac{−\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({tan}\left({x}\right)\right){dx}\overset{???} {=}\frac{\pi}{{e}}{sinh}\left(\mathrm{1}\right) \\ $$

Question Number 138521    Answers: 2   Comments: 0

.......Advanced ... ... ... calculus........ Ω=∫_0 ^( ∞) ((x^2 e^x )/((1+e^x )^3 )) dx =? .....∗∗∗∗∗∗∗∗.....

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{Advanced}\:...\:...\:...\:{calculus}........ \\ $$$$\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} {e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }\:{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:.....\ast\ast\ast\ast\ast\ast\ast\ast..... \\ $$

Question Number 138519    Answers: 0   Comments: 2

Question Number 138516    Answers: 0   Comments: 0

...........calculus ... ... ... (I)......... 𝛗=∫_0 ^( ∞) ((xe^x )/((1+e^x )^3 ))dx =? =(1/2)∫_0 ^( ∞) x.d(((−1)/((1+e^x )^2 )) ) =(1/2)[((−x)/((1+e^x )))]_0 ^∞ +(1/2)∫_0 ^( ∞) (dx/((1+e^x )^2 )) =(1/2)∫_0 ^( ∞) (1/(1+e^x ))dx−(1/2)∫_0 ^( ∞) (e^x /((1+e^x )^2 ))dx =−(1/2)[ln(1+e^(−x) )]_0 ^∞ −(1/2)∫_0 ^( ∞) d(((−1)/(1+e^x ))) =(1/2)ln(2)−(1/2)[((−1)/(1+e^x ))]_0 ^∞ =−(1/2)+(1/2)ln(2) ...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........{calculus}\:...\:...\:...\:\left(\mathrm{I}\right)......... \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{xe}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }{dx}\:=? \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} {x}.{d}\left(\frac{−\mathrm{1}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−{x}}{\left(\mathrm{1}+{e}^{{x}} \right)}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }{dx} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{e}^{−{x}} \right)\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} {d}\left(\frac{−\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right]_{\mathrm{0}} ^{\infty} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:... \\ $$

Question Number 138484    Answers: 0   Comments: 2

Question Number 138485    Answers: 2   Comments: 1

Question Number 138477    Answers: 1   Comments: 0

Question Number 138474    Answers: 0   Comments: 0

hi ! find x such that : ∀ y ∈ [0,1], x ≥ y ⇒ x ≥ 2y.

$$\boldsymbol{\mathrm{hi}}\:!\: \\ $$$$\boldsymbol{\mathrm{find}}\:{x}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\::\:\forall\:{y}\:\in\:\left[\mathrm{0},\mathrm{1}\right],\:{x}\:\geqslant\:{y}\:\Rightarrow\:{x}\:\geqslant\:\mathrm{2}{y}. \\ $$

Question Number 138473    Answers: 1   Comments: 0

hi ! calculate : lim_(x→∞) [((ln(x+1))/(ln (x)))]^(x ln (x)) .

$$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\:\underset{{x}\rightarrow\infty} {{lim}}\:\left[\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\:\left({x}\right)}\right]^{{x}\:{ln}\:\left({x}\right)} . \\ $$

Question Number 138460    Answers: 0   Comments: 0

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