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Question Number 186835 Answers: 0 Comments: 1

Question Number 186833 Answers: 1 Comments: 1

Question Number 186830 Answers: 1 Comments: 1

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Question Number 188822 Answers: 1 Comments: 3

is it a polynomial? p(x)=2x^2 +4x^x −10

$${is}\:{it}\:{a}\:{polynomial}? \\ $$$${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}^{{x}} −\mathrm{10} \\ $$

Question Number 186809 Answers: 1 Comments: 0

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Question Number 186800 Answers: 0 Comments: 0

show that ∫_0 ^( ∞ ) ((tan^(−1) 𝛂x tan^(−1) 𝛃x)/x^2 )dx = (𝛑/2)log{(((𝛂+𝛃)^(𝛂+𝛃) )/(𝛂^𝛂 𝛃^𝛃 ))}

$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty\:} \frac{\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\alpha{x}}\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\beta{x}}}{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{log}\left\{\frac{\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }{\boldsymbol{\alpha}^{\boldsymbol{\alpha}} \boldsymbol{\beta}^{\boldsymbol{\beta}} }\right\}\: \\ $$

Question Number 186790 Answers: 0 Comments: 0

Solve using Fourier′s series: x^2 (1−x)y′′−x(1+x)y′+y=0

$${Solve}\:{using}\:{Fourier}'{s}\:{series}: \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){y}''−{x}\left(\mathrm{1}+{x}\right){y}'+{y}=\mathrm{0} \\ $$

Question Number 186788 Answers: 1 Comments: 0

Question Number 186787 Answers: 0 Comments: 0

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Question Number 186782 Answers: 3 Comments: 0

Question Number 186772 Answers: 0 Comments: 0

Question Number 186771 Answers: 2 Comments: 0

Q : Find the value of the following integral. I = ∫_0 ^( (( π)/( 2))) (( 1)/( 1 + sin^( 4) ( x ) + cos^( 4) ( x ) )) dx = ?

$$ \\ $$$$\:\:\:\:\mathrm{Q}\::\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral}.\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\frac{\:\pi}{\:\mathrm{2}}} \:\frac{\:\:\mathrm{1}}{\:\mathrm{1}\:+\:\mathrm{sin}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:+\:\mathrm{cos}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:}\:\mathrm{d}{x}\:=\:\:?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 186780 Answers: 1 Comments: 0

∫ ((1 + sin x + cos x)/(1 + sin x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\int\:\:\:\frac{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\:\boldsymbol{{x}}}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{{x}}}\:\:\boldsymbol{{dx}}\:\: \\ $$$$ \\ $$

Question Number 186769 Answers: 0 Comments: 0

Given function f : R → R satisfy that (f ○ f)(x) + x = (x+1) f(x) . Find the value of f(1) .

$$\mathrm{Given}\:\:\mathrm{function}\:\:\:{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\:\:\mathrm{satisfy}\:\:\mathrm{that} \\ $$$$\:\:\:\:\left({f}\:\circ\:{f}\right)\left({x}\right)\:+\:{x}\:=\:\left({x}+\mathrm{1}\right)\:{f}\left({x}\right)\:. \\ $$$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\:{f}\left(\mathrm{1}\right)\:. \\ $$

Question Number 186764 Answers: 0 Comments: 3

Question Number 186762 Answers: 3 Comments: 4

Question Number 186758 Answers: 2 Comments: 0

If [t] denotes the integral part of t, then lim_(x→1) [x sin πx] (A) equals 1 (B) equals −1 (C) equals 0 (D) does not exist

$$\mathrm{If}\:\left[{t}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:{t},\:\mathrm{then}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[{x}\:\mathrm{sin}\:\pi{x}\right] \\ $$$$\left(\mathrm{A}\right)\:\:\mathrm{equals}\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\:\mathrm{equals}\:−\mathrm{1} \\ $$$$\left(\mathrm{C}\right)\:\:\mathrm{equals}\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$

Question Number 186752 Answers: 1 Comments: 0

lim_(x→+∞) (((√(x^3 −3x^2 +7))+((x^4 +3))^(1/3) )/( ((x^6 +2x^5 +1))^(1/4) −((x^7 +2x^3 +3))^(1/5) )) Please show work.

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} +\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +\mathrm{2}{x}^{\mathrm{5}} +\mathrm{1}}−\sqrt[{\mathrm{5}}]{{x}^{\mathrm{7}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}}} \\ $$$${Please}\:{show}\:{work}. \\ $$

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