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Question Number 152910 Answers: 0 Comments: 0

∫^ x^x^x dx=

Question Number 152907 Answers: 1 Comments: 0

Find a closed form: Ω=(∫_( 0) ^( 1) ((x^(29) −x^9 )/(x^(40) +1)) dx)(∫_( 0) ^( 1) ((x^(29) −2x^9 )/(x^(40) +4))dx)

$$\mathrm{Find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}: \\ $$$$\Omega=\left(\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{29}} −\mathrm{x}^{\mathrm{9}} }{\mathrm{x}^{\mathrm{40}} +\mathrm{1}}\:\mathrm{dx}\right)\left(\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{29}} −\mathrm{2x}^{\mathrm{9}} }{\mathrm{x}^{\mathrm{40}} +\mathrm{4}}\mathrm{dx}\right) \\ $$

Question Number 152901 Answers: 0 Comments: 0

Question Number 152900 Answers: 1 Comments: 0

Question Number 152899 Answers: 0 Comments: 0

Question Number 152898 Answers: 1 Comments: 2

Question Number 153130 Answers: 1 Comments: 0

if x^5 +x^4 +x^3 +2x^2 +x+1=0 find x^3 - (1/x^3 ) = ?

$$\mathrm{if}\:\:\:\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{3}} \:-\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:? \\ $$

Question Number 152892 Answers: 0 Comments: 2

Solve for real numbers the following system of equations: { ((x^2 - yz = 3)),((y^2 - xz = 1)),((z^2 - xy = - 1)) :}

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{yz}\:=\:\mathrm{3}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{xz}\:=\:\mathrm{1}}\\{\mathrm{z}^{\mathrm{2}} \:-\:\mathrm{xy}\:=\:-\:\mathrm{1}}\end{cases} \\ $$

Question Number 152889 Answers: 0 Comments: 7

Σ_(k=1) ^n ((Σ_(k=1) ^n k^α )/((n+1)^α Σ_(k=1) ^n (1+nα)))=(1/(6o)) α=? α=?q

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\alpha} }{\left({n}+\mathrm{1}\right)^{\alpha} \:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}+{n}\alpha\right)}=\frac{\mathrm{1}}{\mathrm{6}{o}}\: \\ $$$$\alpha=? \\ $$$$\alpha=?{q} \\ $$

Question Number 152887 Answers: 1 Comments: 0

∫_1 ^( 2) (3/( (√((x^2 +3)^3 ))))

$$\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\:\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{3}} }} \\ $$

Question Number 154210 Answers: 0 Comments: 2

Question Number 152881 Answers: 0 Comments: 0

∫_( 0) ^( 1) Li_2 ((x/(1 - x))) log(x) log(1 - x) dx = ?

$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{Li}_{\mathrm{2}} \:\left(\frac{\mathrm{x}}{\mathrm{1}\:-\:\mathrm{x}}\right)\:\mathrm{log}\left(\mathrm{x}\right)\:\mathrm{log}\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 152879 Answers: 0 Comments: 3

lim_(x→+oo) ((Σ_(k=1) ^n k^α )/((n+1)^α Σ_(k=1) ^n (nα+1)))=(1/(6o)) .α=?

$$\underset{{x}\rightarrow+{oo}} {\mathrm{lim}}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\alpha} \:\:\:}{\left({n}+\mathrm{1}\right)^{\alpha} \:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}\alpha+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}{o}}\:\:.\alpha=? \\ $$

Question Number 152904 Answers: 5 Comments: 0

Question Number 152903 Answers: 5 Comments: 0

Question Number 152874 Answers: 2 Comments: 0

solve: I := ∫_0 ^( ∞) ((( tanh (x) )/x) )^( 2) dx = ? m.n.

$$ \\ $$$$\:\:\:{solve}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\:{tanh}\:\left({x}\right)\:}{{x}}\:\right)^{\:\mathrm{2}} {dx}\:=\:? \\ $$$$\:\:{m}.{n}. \\ $$

Question Number 152863 Answers: 1 Comments: 1

∫_(−∞) ^( ∞) (1/( (√(x^2 +1)))) dx

$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{−\infty} ^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\:{dx} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 152861 Answers: 0 Comments: 0

∫_(−∞) ^( ∞) (((ln((x^(√(x^2 +1)) +1)^2 +1))^(−ln(x^2 +1)) )/( (√(x^(∣⌊x⌋∣) +1)))) dx

$$\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\int_{−\infty} ^{\:\infty} \frac{\left(\mathrm{ln}\left(\left({x}^{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)\right)^{−\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} }{\:\sqrt{{x}^{\mid\lfloor{x}\rfloor\mid} +\mathrm{1}}}\:\:{dx} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 152866 Answers: 0 Comments: 0

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

$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{−\infty} ^{\:\infty} \:\frac{\mathrm{ln}\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)}{\left(\mathrm{ln}\left(\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{3}} \right)\right)^{\mathrm{2}} }\:\:{dx} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 152857 Answers: 0 Comments: 0

An electric current passes through two voltmeters in series containing copper sulphate (CuSO₄) and silver nitrate (AgNO₃) respectively. What is the mass of silver deposited in a given time, if the mass of copper deposited in that time is 1g. (Cu = 63 , Ag = 108 valency of Cu is 2 and valency of Ag is 1).

Question Number 152848 Answers: 0 Comments: 5

Question Number 152845 Answers: 0 Comments: 0

Question Number 152844 Answers: 1 Comments: 0

simplify ((3^x + 63)/(21^(x−2) + 7^(x−1) ))

$$\:\:\:\:\:\:\:\:\:\:\:{simplify}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3}^{{x}} \:+\:\mathrm{63}}{\mathrm{21}^{{x}−\mathrm{2}} \:+\:\mathrm{7}^{{x}−\mathrm{1}} }\: \\ $$

Question Number 152841 Answers: 2 Comments: 0

Question Number 152840 Answers: 1 Comments: 0

Question Number 152839 Answers: 0 Comments: 0

Prove that : Ω=∫_0 ^( 1) (( ln^( 3) (1 + x ))/x^( 2) )dx = (3/4) ζ (3 )− 2ln^( 3) ( 2 ) ■ Prepared by: M.N

$$ \\ $$$$\:\:\:\mathrm{Prove}\:\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}^{\:\mathrm{3}} \:\left(\mathrm{1}\:+\:{x}\:\right)}{{x}^{\:\mathrm{2}} }{dx}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\zeta\:\left(\mathrm{3}\:\right)−\:\mathrm{2ln}^{\:\mathrm{3}} \left(\:\mathrm{2}\:\right)\:\:\:\:\:\:\:\:\blacksquare \\ $$$$\:\:\:\:\:\:\:\mathrm{Prepared}\:\mathrm{by}:\:\:\:\:\:\:\mathrm{M}.\mathrm{N} \\ $$$$ \\ $$

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