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Question Number 201011    Answers: 0   Comments: 0

Prove that ∫_0 ^∞ ((2arctan((t/x)))/(e^(2𝛑t) βˆ’1))dt=Inπšͺ(x)βˆ’xIn(x)+xβˆ’(1/2)In(((2𝛑)/x)) Michael faraday

$$\boldsymbol{{Prove}}\:\boldsymbol{{that}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{t}}}{\boldsymbol{{x}}}\right)}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi{t}}} βˆ’\mathrm{1}}\boldsymbol{{dt}}=\boldsymbol{{In}\Gamma}\left(\boldsymbol{{x}}\right)βˆ’\boldsymbol{{xIn}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{x}}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{In}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\boldsymbol{{x}}}\right) \\ $$$$\boldsymbol{{Michael}}\:\boldsymbol{{faraday}} \\ $$

Question Number 201008    Answers: 1   Comments: 0

Question Number 201004    Answers: 2   Comments: 0

Question Number 200984    Answers: 3   Comments: 0

Question Number 200980    Answers: 1   Comments: 0

Question Number 200979    Answers: 1   Comments: 0

Question Number 200978    Answers: 3   Comments: 0

Question Number 200968    Answers: 0   Comments: 0

Question Number 200971    Answers: 1   Comments: 0

Question Number 200976    Answers: 4   Comments: 0

Question Number 200973    Answers: 0   Comments: 0

Question Number 200972    Answers: 1   Comments: 0

Question Number 200970    Answers: 3   Comments: 0

Question Number 200964    Answers: 1   Comments: 0

Question Number 200961    Answers: 2   Comments: 0

βˆ’x^3 +1=((βˆ’x+1))^(1/3) β‡’x=?

$$\: \\ $$$$\:\:\:\:\:βˆ’{x}^{\mathrm{3}} +\mathrm{1}=\sqrt[{\mathrm{3}}]{βˆ’{x}+\mathrm{1}}\:\Rightarrow{x}=? \\ $$$$ \\ $$

Question Number 200960    Answers: 3   Comments: 0

Question Number 200958    Answers: 2   Comments: 0

Question Number 200954    Answers: 0   Comments: 1

Question Number 200937    Answers: 2   Comments: 0

Question Number 200936    Answers: 0   Comments: 1

Question Number 200934    Answers: 1   Comments: 0

Question Number 200933    Answers: 0   Comments: 0

∫coth (ln [(√(tanh (ln ((√(sec^(βˆ’1) (x)^(1/4) )))))) ])

$$ \\ $$$$\int\mathrm{coth}\:\left(\mathrm{ln}\:\left[\sqrt{\mathrm{tanh}\:\left(\mathrm{ln}\:\left(\sqrt{\mathrm{sec}^{βˆ’\mathrm{1}} \:\:\sqrt[{\mathrm{4}}]{{x}}\:\:}\right)\right)}\:\right]\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 201000    Answers: 0   Comments: 0

find 1)(0,2) βˆͺ {3} 2)[0,2] βˆͺ {3} 3) (βˆ’5,5) βˆͺ {6} 4) (βˆ’2 , 2 ) βˆͺ [3,5]

$${find}\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\left(\mathrm{0},\mathrm{2}\right)\:\cup\:\left\{\mathrm{3}\right\}\:\: \\ $$$$ \\ $$$$\left.\:\:\mathrm{2}\right)\left[\mathrm{0},\mathrm{2}\right]\:\cup\:\left\{\mathrm{3}\right\} \\ $$$$ \\ $$$$\left.\:\:\mathrm{3}\right)\:\left(βˆ’\mathrm{5},\mathrm{5}\right)\:\cup\:\left\{\mathrm{6}\right\} \\ $$$$\:\: \\ $$$$\left.\:\:\mathrm{4}\right)\:\left(βˆ’\mathrm{2}\:,\:\mathrm{2}\:\right)\:\cup\:\left[\mathrm{3},\mathrm{5}\right] \\ $$

Question Number 200942    Answers: 1   Comments: 0

Question Number 200929    Answers: 0   Comments: 2

A ball lies on the function z=xy at the point (1,2,2). Find the point in the xyβˆ’plane where the ball will touch it. Calculus 2 problem.

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:{xy}βˆ’\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{touch}\:\mathrm{it}. \\ $$$$\mathrm{Calculus}\:\mathrm{2}\:\mathrm{problem}. \\ $$

Question Number 200925    Answers: 1   Comments: 0

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