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Question Number 69641    Answers: 0   Comments: 0

Solve arctg(1βˆ’x) + (1/(arcctg(1+x))) = (𝛑/4)

$$\boldsymbol{{Solve}}\:\:\boldsymbol{{arctg}}\left(\mathrm{1}βˆ’\boldsymbol{{x}}\right)\:+\:\frac{\mathrm{1}}{\boldsymbol{{arcctg}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$

Question Number 69637    Answers: 1   Comments: 0

...now try this one: ∫(dx/(x^(1/2) βˆ’x^(1/3) βˆ’x^(1/6) ))=

$$...\mathrm{now}\:\mathrm{try}\:\mathrm{this}\:\mathrm{one}: \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{2}} βˆ’{x}^{\mathrm{1}/\mathrm{3}} βˆ’{x}^{\mathrm{1}/\mathrm{6}} }= \\ $$

Question Number 69623    Answers: 1   Comments: 0

∫(1/((√x) + (x)^(1/3) )) dx

$$\int\frac{\mathrm{1}}{\sqrt{{x}}\:+\:\sqrt[{\mathrm{3}}]{{x}}}\:{dx} \\ $$

Question Number 69616    Answers: 0   Comments: 4

Question Number 69606    Answers: 0   Comments: 0

Question Number 69603    Answers: 1   Comments: 0

∫ x^3 arcsinxdx

$$\int\:{x}^{\mathrm{3}} {arcsinxdx} \\ $$

Question Number 69597    Answers: 1   Comments: 1

Question Number 69594    Answers: 1   Comments: 0

Question Number 69593    Answers: 1   Comments: 2

Question Number 69576    Answers: 1   Comments: 1

∫_(βˆ’2) ^( 2) (x^3 cos(x/2)+(1/2))(√(4βˆ’x^2 ))dx

$$\int_{βˆ’\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}βˆ’{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 69574    Answers: 1   Comments: 0

Question Number 69573    Answers: 0   Comments: 1

Question Number 69572    Answers: 1   Comments: 0

Question Number 69571    Answers: 1   Comments: 0

Question Number 69570    Answers: 0   Comments: 0

Question Number 69569    Answers: 2   Comments: 0

Question Number 69568    Answers: 1   Comments: 3

Question Number 69567    Answers: 0   Comments: 2

Question Number 69566    Answers: 1   Comments: 0

Question Number 69565    Answers: 0   Comments: 0

Question Number 69564    Answers: 0   Comments: 3

let f(a) =∫_0 ^∞ (dx/(x^4 βˆ’2x^2 +a)) with a real and a>1 1) determine a explicit form for f(a) 2) calculate g(a) =∫_0 ^∞ (dx/((x^4 βˆ’2x^2 +a)^2 )) 3) find the values of integrals ∫_0 ^∞ (dx/(x^4 βˆ’2x^2 +3)) and ∫_0 ^∞ (dx/((x^4 βˆ’2x^2 +3)^2 ))

$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} βˆ’\mathrm{2}{x}^{\mathrm{2}} \:+{a}}\:\:\:{with}\:{a}\:{real}\:{and}\:{a}>\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} βˆ’\mathrm{2}{x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} βˆ’\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} βˆ’\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Question Number 69563    Answers: 0   Comments: 1

calculate ∫_0 ^∞ (dx/(x^4 βˆ’x^2 +1))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} βˆ’{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$

Question Number 69610    Answers: 0   Comments: 0

Question Number 69609    Answers: 0   Comments: 2

Question Number 69608    Answers: 1   Comments: 2

Question Number 69607    Answers: 0   Comments: 4

without using lhospital please prove that lim_(xβ†’0) ((xβˆ’sin x)/x^3 ) = (1/6) I want every method possible because someone challenge me

$$\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{lhospital}}\:\boldsymbol{{please}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}βˆ’\boldsymbol{{sin}}\:\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{want}}\:\boldsymbol{{every}}\:\boldsymbol{{method}} \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{because}}\:\boldsymbol{{someone}} \\ $$$$\boldsymbol{{challenge}}\:\boldsymbol{{me}}\: \\ $$

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