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Question Number 90003 Answers: 0 Comments: 0

show that ∫_(−∞) ^∞ (dx/(1+(x+tan(x))^(2 ) ))=π

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

Question Number 89994 Answers: 2 Comments: 0

x−y=3(√(xy)) ((x/y)−1)^3 +((y/x)−1)^3 = ?

$$\mathrm{x}−\mathrm{y}=\mathrm{3}\sqrt{\mathrm{xy}} \\ $$$$\left(\frac{\mathrm{x}}{\mathrm{y}}−\mathrm{1}\right)^{\mathrm{3}} +\left(\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{1}\right)^{\mathrm{3}} \:=\:? \\ $$

Question Number 89991 Answers: 0 Comments: 2

if a,b > 0 and (a^2 /b^2 ) = (5/3) find ((a^2 +b^2 )/(ab))

$$\mathrm{if}\:\mathrm{a},\mathrm{b}\:>\:\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{find}\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{ab}} \\ $$

Question Number 89986 Answers: 0 Comments: 3

∫_0 ^1 (−1)^(⌊(1/x)⌋) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{\lfloor\frac{\mathrm{1}}{{x}}\rfloor} \:{dx} \\ $$

Question Number 89980 Answers: 0 Comments: 2

Question Number 89978 Answers: 0 Comments: 1

(x (dy/dx)−y)(cos (((2y)/x))) = −3x^4

$$\left(\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{y}\right)\left(\mathrm{cos}\:\left(\frac{\mathrm{2y}}{\mathrm{x}}\right)\right)\:=\:−\mathrm{3x}^{\mathrm{4}} \\ $$

Question Number 89977 Answers: 0 Comments: 2

Question Number 89973 Answers: 1 Comments: 1

x (dy/dx) −y = x^2 tan ((y/x))

$${x}\:\frac{{dy}}{{dx}}\:−{y}\:=\:{x}^{\mathrm{2}} \:\mathrm{tan}\:\left(\frac{{y}}{{x}}\right)\: \\ $$

Question Number 89970 Answers: 0 Comments: 1

xy (dy/dx) = y^2 ((x^3 /(x^2 +1)))

$$\mathrm{xy}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}^{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\: \\ $$

Question Number 89955 Answers: 1 Comments: 2

9^(x+1) ∤28(3^x )+3=0

$$\mathrm{9}^{\mathrm{x}+\mathrm{1}} \nmid\mathrm{28}\left(\mathrm{3}^{\mathrm{x}} \right)+\mathrm{3}=\mathrm{0} \\ $$

Question Number 89953 Answers: 0 Comments: 1

solvethefollowingequation 5^(2x+y) =625and2^(4x∤2y) =(1/6)

$$\mathrm{solvethefollowingequation} \\ $$$$\mathrm{5}^{\mathrm{2x}+\mathrm{y}} =\mathrm{625and2}^{\mathrm{4x}\nmid\mathrm{2y}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Question Number 89951 Answers: 0 Comments: 1

log _2 (sin (x+((5π)/(12)))) + log _2 (sin (x+(π/(12))))=−1

$$\mathrm{log}\:_{\mathrm{2}} \:\left(\mathrm{sin}\:\left({x}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right)\:+\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right)\right)=−\mathrm{1} \\ $$

Question Number 89950 Answers: 0 Comments: 1

Question Number 89946 Answers: 1 Comments: 0

∫ _(−(π/2)) ^(π/2) (dx/(1+e^(sin x) ))

$$\int\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} } \\ $$

Question Number 89958 Answers: 0 Comments: 1

prove that (1+sin x/1+cos 3(1sin x/1+cosec x)=tanx

$${prove}\:{that}\:\left(\mathrm{1}+\mathrm{sin}\:{x}/\mathrm{1}+\mathrm{cos}\:\mathrm{3}\left(\mathrm{1sin}\:{x}/\mathrm{1}+\mathrm{cosec}\:{x}\right)={tanx}\right. \\ $$

Question Number 89956 Answers: 0 Comments: 1

simplifyκgivingκyourκanswerκinκindexκform (√((ac^2 )/(9a^2 c^4 )))

$$\mathrm{simplify}\kappa\mathrm{giving}\kappa\mathrm{your}\kappa\mathrm{answer}\kappa\mathrm{in}\kappa\mathrm{index}\kappa\mathrm{form} \\ $$$$\sqrt{\frac{\mathrm{ac}^{\mathrm{2}} }{\mathrm{9a}^{\mathrm{2}} \mathrm{c}^{\mathrm{4}} }} \\ $$

Question Number 89938 Answers: 0 Comments: 1

If x(x+1) = 1 find (x+1)^3 −(1/((x+1)^3 ))

$$\mathrm{If}\:\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{1}\: \\ $$$$\mathrm{find}\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 89937 Answers: 0 Comments: 1

Prove that for all complex such as ∣z∣<1= Σ_(n=1) ^∞ (z^n /((z^n −1)^2 )) +Σ_(n=1) ^∞ ((nz^n )/(z^n −1)) = 0

$${Prove}\:{that}\:{for}\:{all}\:{complex}\:{such}\:{as}\:\mid{z}\mid<\mathrm{1}= \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\left({z}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{nz}^{{n}} }{{z}^{{n}} −\mathrm{1}}\:=\:\mathrm{0}\: \\ $$

Question Number 89936 Answers: 1 Comments: 0

Prove that Σ_(p≥1,q≥1) (1/(pq(p+q−1))) =(π^2 /3)

$${Prove}\:{that}\:\underset{{p}\geqslant\mathrm{1},{q}\geqslant\mathrm{1}} {\sum}\:\:\frac{\mathrm{1}}{{pq}\left({p}+{q}−\mathrm{1}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\: \\ $$

Question Number 89934 Answers: 0 Comments: 0

Let x∈]0;1[ Prove that Σ_(n=1) ^∞ (x^n /(1+x^n )) +Σ_(n=1) ^∞ (((−x)^n )/(1−x^n )) = 0

$$\left.{Let}\:{x}\in\right]\mathrm{0};\mathrm{1}\left[\:\:{Prove}\:{that}\right. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{\mathrm{1}+{x}^{{n}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} }{\mathrm{1}−{x}^{{n}} }\:=\:\mathrm{0} \\ $$

Question Number 89928 Answers: 1 Comments: 0

Question Number 89925 Answers: 0 Comments: 0

x^2 (yy′′−y^2 )+xyy′ = y(√(x^2 (y′)^2 +y^2 ))

$${x}^{\mathrm{2}} \left({yy}''−{y}^{\mathrm{2}} \right)+{xyy}'\:=\:{y}\sqrt{{x}^{\mathrm{2}} \left({y}'\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }\: \\ $$

Question Number 89922 Answers: 0 Comments: 2

Question Number 89918 Answers: 0 Comments: 1

∫ ((x tan^(−1) (x))/((1+x^2 )^(3/2) )) dx

$$\int\:\frac{\mathrm{x}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\mathrm{dx}\: \\ $$

Question Number 89913 Answers: 1 Comments: 1

Question Number 89908 Answers: 0 Comments: 6

Solve the differential equstion: (d^2 y/dx^2 ) = ((y_0 − 2y_(−1) + y_(−2) )/h^2 )

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equstion}: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:=\:\:\:\frac{\mathrm{y}_{\mathrm{0}} \:\:−\:\:\mathrm{2y}_{−\mathrm{1}} \:\:+\:\:\mathrm{y}_{−\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} } \\ $$

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