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Question Number 220245 Answers: 1 Comments: 0

Question Number 220244 Answers: 3 Comments: 0

Question Number 220243 Answers: 5 Comments: 0

Question Number 220242 Answers: 0 Comments: 0

∫ ((ln x)/((1 + x^2 )^2 )) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{{ln}\:{x}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:^{\mathrm{2}} }\:\:{dx} \\ $$$$ \\ $$

Question Number 220232 Answers: 1 Comments: 0

prove that (π/(16)) < ∫_0 ^( 1 ) (√((x(1−x))/(sin(πx)+cos(πx)+2))) dx<(π/8)

$$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\:\frac{\pi}{\mathrm{16}}\:<\:\int_{\mathrm{0}} ^{\:\mathrm{1}\:} \sqrt{\frac{{x}\left(\mathrm{1}−{x}\right)}{{sin}\left(\pi{x}\right)+{cos}\left(\pi{x}\right)+\mathrm{2}}}\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\: \\ $$

Question Number 220231 Answers: 4 Comments: 0

Question Number 220224 Answers: 5 Comments: 0

calcul together of definition of and calcul the derive f^ ^′ f(x)= x(√((x−1)/(x+1)))

$${calcul}\:{together}\:{of}\:{definition}\:{of}\:{and} \\ $$$${calcul}\:{the}\:{derive}\:{f}^{\:} \:^{'} \\ $$$${f}\left({x}\right)=\:\:{x}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$

Question Number 220221 Answers: 0 Comments: 2

Q.The density of an object of mass M is δ and the density of the air is ρ. the mass of of the object is measured with the help of a metal weight of mass m . the density of the metal weight is d. if ρ≪δ them show that the real mass M will be m(1−(ρ/d) )(1+(ρ/δ)) I have managed to M=((m(1−(ρ/d)))/((1−(ρ/δ)))) but I can not figure it to the end please help

$${Q}.{The}\:{density}\:{of}\:{an}\:{object}\:{of}\:{mass}\:{M}\:{is}\:\delta\:{and}\:{the}\:{density}\:{of}\:{the}\:{air}\:{is}\:\rho. \\ $$$${the}\:{mass}\:{of}\:{of}\:{the}\:{object}\:{is}\:{measured}\:{with}\:\:{the}\:{help}\:{of}\:{a}\:{metal}\:{weight}\:{of}\:{mass}\:{m}\:. \\ $$$${the}\:{density}\:{of}\:{the}\:{metal}\:{weight}\:{is}\:{d}. \\ $$$${if}\:\rho\ll\delta\:{them}\:{show}\:{that}\:{the}\:{real}\:{mass}\:{M}\:{will}\:{be} \\ $$$${m}\left(\mathrm{1}−\frac{\rho}{{d}}\:\right)\left(\mathrm{1}+\frac{\rho}{\delta}\right) \\ $$$${I}\:{have}\:{managed}\:{to}\:{M}=\frac{{m}\left(\mathrm{1}−\frac{\rho}{{d}}\right)}{\left(\mathrm{1}−\frac{\rho}{\delta}\right)} \\ $$$${but}\:{I}\:{can}\:{not}\:{figure}\:{it}\:{to}\:{the}\:{end} \\ $$$${please}\:{help} \\ $$

Question Number 220208 Answers: 3 Comments: 2

Question Number 220200 Answers: 0 Comments: 0

∫_1 ^( α) (((x − 1)^n )/(e^x − x − 1))dx

$$ \\ $$$$\int_{\mathrm{1}} ^{\:\alpha} \frac{\left({x}\:−\:\mathrm{1}\right)^{{n}} }{{e}^{{x}} \:−\:{x}\:−\:\mathrm{1}}{dx} \\ $$

Question Number 220193 Answers: 3 Comments: 0

x^8 =21x+13 ; x∈R x=?

$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{8}} =\mathrm{21}\boldsymbol{{x}}+\mathrm{13}\:\:\:\:\:\:\:\:\:\:;\:\:\:\:\boldsymbol{{x}}\in{R} \\ $$$$\:\:\:\:\boldsymbol{{x}}=? \\ $$

Question Number 220192 Answers: 0 Comments: 0

Question Number 220190 Answers: 1 Comments: 0

i^i =e^(−(π/2)) and we can renote complex number i as ((0,(−1)),(1,( 0)) ) i^i = ((0,(−1)),(1,( 0)) )^ ((0,(−1)),(1,( 0)) ) But why Matrix Exponent Calculate Dosen′t defined?? I mean A,B∈mat(m,m) why A^B dosen′t defined??

$${i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \: \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\mathrm{complex}\:\mathrm{number}\:\boldsymbol{{i}}\:\mathrm{as}\:\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\boldsymbol{{i}}^{\boldsymbol{{i}}} =\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}^{\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}} \: \\ $$$$\:\mathrm{But}\:\mathrm{why}\:\mathrm{Matrix}\:\mathrm{Exponent}\:\mathrm{Calculate}\:\mathrm{Dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$$$\:\mathrm{I}\:\mathrm{mean}\:{A},{B}\in\mathrm{mat}\left({m},{m}\right) \\ $$$$\mathrm{why}\:\mathrm{A}^{\mathrm{B}} \:\mathrm{dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$

Question Number 220184 Answers: 2 Comments: 0

∫_0 ^(π/(12)) (√((sec^4 α+5sec^5 αsin α)/((2−sec^2 α)(125tan^3 α+25tan^2 α+5tan α+1))))dα

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \sqrt{\frac{\mathrm{sec}^{\mathrm{4}} \alpha+\mathrm{5sec}^{\mathrm{5}} \alpha\mathrm{sin}\:\alpha}{\left(\mathrm{2}−\mathrm{sec}^{\mathrm{2}} \alpha\right)\left(\mathrm{125tan}^{\mathrm{3}} \alpha+\mathrm{25tan}^{\mathrm{2}} \alpha+\mathrm{5tan}\:\alpha+\mathrm{1}\right)}}{d}\alpha \\ $$

Question Number 220323 Answers: 0 Comments: 0

Q1. 𝛀;={(x,y);x^2 +y^2 ≤1} (1/2) ∮_( ∂𝛀) x∙dy−y∙dy=?? Q2. S; R^2 →R^3 S(u,v)=rsin(u)cos(v)e_1 ^→ +rsin(u)sin(v)e_2 ^→ +rcos(u)e_3 ^→ F^→ ;R^3 →R^3 F^→ (x,y,z)=−(x/( (√(x^2 +y^2 +z^2 ))))e_1 ^→ −(y/( (√(x^2 +y^2 +z^2 ))))e_2 ^→ −(z/( (√(x^2 +y^2 +z^2 ))))e_3 ^→ ∫∫_( D) F^→ ∙dS^→ =∫∫∫_( K) ▽^→ ∙F^→ dV=??? Q3. if ∮_( C) F^→ ∙dl=0 Prove ▽^→ ×F^→ =0 Q4. Prove if ▽^→ ×F^→ ≠0 → ∮_( C) F^→ ∙dl≠0

Question Number 220179 Answers: 1 Comments: 0

∫ (ds/( (√(s^2 +1))(s+(√(s^2 +1)))^(−ν) ))

$$\int\:\:\:\frac{\mathrm{d}{s}}{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} } \\ $$

Question Number 220178 Answers: 1 Comments: 0

evaluate ∫_w ^( ∞) ((e^s ∙𝚪(0,s))/s)ds

$$\mathrm{evaluate} \\ $$$$\int_{{w}} ^{\:\infty} \:\frac{{e}^{{s}} \centerdot\boldsymbol{\Gamma}\left(\mathrm{0},{s}\right)}{{s}}\mathrm{d}{s} \\ $$

Question Number 220177 Answers: 1 Comments: 0

∫_(−1) ^( 0) cos(((ln(z+1))/z)) dz

$$\int_{−\mathrm{1}} ^{\:\mathrm{0}} \:\:\mathrm{cos}\left(\frac{\mathrm{ln}\left({z}+\mathrm{1}\right)}{{z}}\right)\:\mathrm{d}{z} \\ $$

Question Number 220176 Answers: 0 Comments: 0

∫ (dx/(1 + x^7 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{7}} }\: \\ $$$$ \\ $$

Question Number 220164 Answers: 2 Comments: 0

if α^2 −5α+2=0 & β^2 −5β+2=0 then ((4α+β^5 )/(5β^2 ))=?

$${if}\:\:\alpha^{\mathrm{2}} −\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:\:\&\:\:\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{2}=\mathrm{0} \\ $$$${then}\:\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=? \\ $$

Question Number 220160 Answers: 0 Comments: 0

for all x , y [0 , 1] ; prove that; [ (((x^3 + y^3 + 𝛇(3)))^(1/(3 )) /(1 + e^(−x^2 y^2 ) )) + (((x^4 + 𝚪(y+1)))^(1/(4 )) /((1 + y^2 )^(1/3) )) + ((ln(1 + x^5 + y^5 ))/( (√(1 + x^2 + y^2 )))) + Li_2 (xy) + ((√(x^6 + y^6 +1 ))/((1 + x^3 y^3 )^(1/2) )) ≤ (e^(xy) /(1 + x + y )) + ((ln (1 + x^2 + y^2 ) ))^(1/(3 )) + ((2𝛇(2))/( (√(1 + x^2 y^2 )))) + ((x^8 + y^8 + 1))^(1/(4 )) ]

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x}\:,\:{y}\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\left[\:\frac{\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{{x}}^{\mathrm{3}} \:+\:\boldsymbol{{y}}^{\mathrm{3}} \:+\:\boldsymbol{\zeta}\left(\mathrm{3}\right)}}{\mathrm{1}\:+\:\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} } \:}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{4}} +\:\boldsymbol{\Gamma}\left(\boldsymbol{{y}}+\mathrm{1}\right)}}{\left(\mathrm{1}\:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:+\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{5}} \:+\:\boldsymbol{{y}}^{\mathrm{5}} \right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\boldsymbol{{xy}}\right)\:+\:\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{6}} \:+\:\boldsymbol{{y}}^{\mathrm{6}} \:+\mathrm{1}\:}}{\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{2}} }\: \\ $$$$\left.\:\:\:\:\:\:\leqslant\:\frac{\boldsymbol{{e}}^{\boldsymbol{{xy}}} }{\mathrm{1}\:+\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:}\:+\:\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{\mathrm{ln}}\:\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)\:}\:+\:\frac{\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} }}\:+\:\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{8}} \:+\:\boldsymbol{{y}}^{\mathrm{8}} \:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\right]\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 220159 Answers: 0 Comments: 1

for all x, y ∈ [0 , 1] ; prove that; (1/( (√(1 + x^4 )))) + (2/( (√(1 + y^4 )))) + (2/( (√(4 + (x + y)^4 )))) + ((2(√2))/( (√(2+ x^2 y^2 + y^3 )))) ≤ (2/( (√(1 + x^2 y^2 )))) + (2/(^4 (√(1 + x^5 + y^5 )))) + ln(e+((x^3 y+y^3 x)/(1 + xy))) + (1/((1+x+y)^3 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}} \:+\:{y}^{\mathrm{5}} }}\:+\:\mathrm{ln}\left({e}+\frac{{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {x}}{\mathrm{1}\:+\:{xy}}\right)\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{y}\right)^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220141 Answers: 0 Comments: 0

Σ_(n=1) ^∞ Σ_(m=1) ^∞ (1/((n^2 +m^2 )^(3/2) ))=?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{m}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=? \\ $$

Question Number 220131 Answers: 2 Comments: 0

If f(x,y)=(((x^2 +y^2 )^n )/(2n(2n−1)))+xφ((y/x))+Ψ((y/x)), then using Euler′s theorem on homogenous functions,show that x^2 ((δ^2 f)/(δx^2 ))+2xy((δ^2 f)/(δxδy))+y^2 ((δ^2 f)/(δy^2 ))=(x^2 +y^2 )^n

$${If}\:\:\:{f}\left({x},{y}\right)=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} }{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}+{x}\phi\left(\frac{{y}}{{x}}\right)+\Psi\left(\frac{{y}}{{x}}\right), \\ $$$${then}\:{using}\:{Euler}'{s}\:{theorem}\:{on}\:{homogenous}\:{functions},{show}\:{that} \\ $$$${x}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{x}^{\mathrm{2}} }+\mathrm{2}{xy}\frac{\delta^{\mathrm{2}} {f}}{\delta{x}\delta{y}}+{y}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{y}^{\mathrm{2}} }=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} \\ $$

Question Number 220121 Answers: 1 Comments: 0

Let H_h =p_(h+1) /p_h , p_h ∈P , p_1 =2 Π_(h=1) ^∞ H_h =?? (Π_(h=1) ^∞ H_h =(3/2)∙(5/3)∙(7/5).........)

$$\mathrm{Let}\:{H}_{{h}} ={p}_{{h}+\mathrm{1}} /{p}_{{h}} \:,\:{p}_{{h}} \in\mathbb{P}\:,\:{p}_{\mathrm{1}} =\mathrm{2} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\prod}}\:{H}_{{h}} =??\:\left(\underset{{h}=\mathrm{1}} {\overset{\infty} {\prod}}\:{H}_{{h}} =\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\mathrm{5}}{\mathrm{3}}\centerdot\frac{\mathrm{7}}{\mathrm{5}}.........\right) \\ $$

Question Number 220321 Answers: 1 Comments: 2

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