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Question Number 85721    Answers: 1   Comments: 0

show that ∫_0 ^∞ ((e^(−x) ln(x))/(√x))dx=−(√π)(γ+ln(4))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\sqrt{{x}}}{dx}=−\sqrt{\pi}\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$

Question Number 85718    Answers: 1   Comments: 0

∫((sin(x)−cos(3x))/(sin(x)−cos(2x)))dx

$$\int\frac{{sin}\left({x}\right)−{cos}\left(\mathrm{3}{x}\right)}{{sin}\left({x}\right)−{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$

Question Number 85717    Answers: 1   Comments: 0

∫_0 ^2 x^4 (√(1−x^2 )) dx ∫_0 ^1 x^(10) (1−x^n )dx

$$\int_{\mathrm{0}} ^{\mathrm{2}} {x}^{\mathrm{4}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{10}} \left(\mathrm{1}−{x}^{{n}} \right){dx} \\ $$$$ \\ $$

Question Number 85711    Answers: 0   Comments: 2

∫_(−4) ^2 ((2x + 1)/((x^2 + x + 1)^(3/2) )) dx

$$\:\underset{−\mathrm{4}} {\overset{\mathrm{2}} {\int}}\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{\left({x}^{\mathrm{2}} +\:{x}\:+\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\:{dx} \\ $$

Question Number 85709    Answers: 1   Comments: 0

Question Number 85708    Answers: 1   Comments: 0

lim_(n−∞) /((U_n +1)/(Un))/ >0 Test for convergence

$${li}\underset{{n}−\infty} {{m}}\:\:/\frac{{U}_{{n}} +\mathrm{1}}{{Un}}/\:\:\:>\mathrm{0} \\ $$$${Test}\:{for}\:{convergence} \\ $$

Question Number 85706    Answers: 1   Comments: 0

Find the general solution of x^2 (√(y^2 +9)) dx + 5 (√(x^2 −3)) y dy = 0

$${Find}\:\:{the}\:\:{general}\:\:{solution}\:\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\:\:{dx}\:+\:\mathrm{5}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}\:\:{y}\:{dy}\:\:=\:\:\mathrm{0} \\ $$

Question Number 85701    Answers: 1   Comments: 3

∫ ((√(3x−1))/(√(2x+1))) dx

$$\int\:\frac{\sqrt{\mathrm{3x}−\mathrm{1}}}{\sqrt{\mathrm{2x}+\mathrm{1}}}\:\mathrm{dx}\: \\ $$

Question Number 85698    Answers: 0   Comments: 0

please any recommendation of a youtube video on General conics??

$$\mathrm{please}\:\mathrm{any}\:\mathrm{recommendation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{youtube}\:\mathrm{video} \\ $$$$\mathrm{on}\:\mathrm{General}\:\mathrm{conics}?? \\ $$

Question Number 85696    Answers: 0   Comments: 0

Montrer que: (√5)+(√(30))+(√(50))<(√(10))+(√(20))+(√(60)) {niveau second)

$$\mathrm{Montrer}\:\mathrm{que}: \\ $$$$\sqrt{\mathrm{5}}+\sqrt{\mathrm{30}}+\sqrt{\mathrm{50}}<\sqrt{\mathrm{10}}+\sqrt{\mathrm{20}}+\sqrt{\mathrm{60}} \\ $$$$\left\{\mathrm{niveau}\:\mathrm{second}\right) \\ $$

Question Number 85695    Answers: 0   Comments: 0

Question Number 85694    Answers: 0   Comments: 1

log_(((x/(x−3)))) (7) < log_(((x/3))) (7)

$$\mathrm{log}_{\left(\frac{{x}}{{x}−\mathrm{3}}\right)} \left(\mathrm{7}\right)\:<\:\mathrm{log}_{\left(\frac{{x}}{\mathrm{3}}\right)} \:\left(\mathrm{7}\right)\: \\ $$

Question Number 85677    Answers: 2   Comments: 3

∫_0 ^π ((sin (((21x)/2)))/(sin ((x/2)))) dx

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx}\: \\ $$

Question Number 85676    Answers: 0   Comments: 15

∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)

$$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{sin}\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{t}\right)^{\mathrm{4}} } \\ $$$$=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\mathrm{4cos}^{\mathrm{4}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)\:{dt} \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}=\:{u}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} {u}\:×\mathrm{2}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\:}}\left(\mathrm{tan}\:^{\mathrm{2}} {u}+\mathrm{1}\right)\:{d}\left(\mathrm{tan}\:{u}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} {u}\:+\:\mathrm{tan}\:{u}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{0}−\left(−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\right]=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Question Number 85670    Answers: 0   Comments: 3

Question Number 85669    Answers: 1   Comments: 4

∫ ((√(1+x))/(√(1−x))) dx

$$\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$$$ \\ $$

Question Number 85668    Answers: 0   Comments: 2

z = 2 + i find arg(z)

$$\mathrm{z}\:=\:\mathrm{2}\:+\:\mathrm{i}\: \\ $$$$\mathrm{find}\:\mathrm{arg}\left(\mathrm{z}\right) \\ $$

Question Number 85667    Answers: 1   Comments: 0

∫ (dx/(√(1−sin 2x)))

$$\int\:\frac{{dx}}{\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}}\: \\ $$

Question Number 85666    Answers: 0   Comments: 0

Question Number 85664    Answers: 0   Comments: 2

Show that a group order 100 is not simple

$$\boldsymbol{{S}\mathrm{how}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{order}}\:\mathrm{100}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{simple}} \\ $$

Question Number 85656    Answers: 0   Comments: 1

Question Number 85655    Answers: 0   Comments: 0

Question Number 85653    Answers: 0   Comments: 1

if f≥0 and (d/dx)(f(x))^2 =(f′(x))^2 and f(0)=1 find f(x) if f(x)=((4x^3 )/(x^2 +1)) find f^( −1) (x) and (f^( −1) )^′ (2)

$${if}\:\:{f}\geqslant\mathrm{0}\:\:{and}\:\:\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)^{\mathrm{2}} =\left({f}'\left({x}\right)\right)^{\mathrm{2}} \:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${find}\:{f}\left({x}\right)\:\:\: \\ $$$$ \\ $$$${if}\:{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:\:{find}\:{f}^{\:−\mathrm{1}} \left({x}\right)\:\:\:{and}\:\left({f}^{\:−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right) \\ $$

Question Number 85649    Answers: 1   Comments: 1

Proove that : ((√(2−(√3)))/2) = (((√6)−(√2))/4)

$$\mathrm{Proove}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Question Number 85648    Answers: 2   Comments: 0

∫(((x^3 −4))/((x+1)))dx

$$\int\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{4}\right)}{\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$

Question Number 85646    Answers: 0   Comments: 0

show that ∫(1/([x(x−1)(x−2)(x−3)...(x−m)]^2 ))dx= =(1/((m!)^2 ))Σ_(n=0) ^m ( ((m),(n) )^2 /(n−x))+(2/((m!)^2 ))ln∣Π_(n=0) ^m (x−n)^( ((m),(n) )^2 (H_(m−n) −H_n )) ∣+c

$${show}\:{that} \\ $$$$\int\frac{\mathrm{1}}{\left[{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)...\left({x}−{m}\right)\right]^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\left({m}!\right)^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} }{{n}−{x}}+\frac{\mathrm{2}}{\left({m}!\right)^{\mathrm{2}} }{ln}\mid\underset{{n}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}−{n}\right)^{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \left({H}_{{m}−{n}} −{H}_{{n}} \right)} \mid+{c} \\ $$

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