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x^( 2) − 4x −1=0 α , β are roots α^( 3) + 17β +5 =? −−−solution−−− α is root ⇒ α^( 2) −4α −1=0 ⇒ α^( 2) = 4α +1 ✓ α^( 3) + 17β +5 = α . α^( 2) +17β +5 = α ( 4α +1 )+ 17β +5 = 4α^( 2) + α + 17β +5 = 4 (4α +1 )+α +17β +5=17(α+β)+9 = 17S +9= 17 (4 )+9=77

$$ \\ $$$$\:\:\:\:{x}^{\:\mathrm{2}} −\:\mathrm{4}{x}\:−\mathrm{1}=\mathrm{0}\:\: \\ $$$$\:\:\:\:\:\alpha\:,\:\beta\:\:{are}\:{roots}\: \\ $$$$\:\:\:\:\:\alpha^{\:\mathrm{3}} \:+\:\mathrm{17}\beta\:+\mathrm{5}\:=? \\ $$$$\:\:−−−{solution}−−− \\ $$$$\:\:\:\alpha\:\:\:{is}\:{root}\:\:\:\Rightarrow\:\alpha^{\:\mathrm{2}} −\mathrm{4}\alpha\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:\alpha^{\:\mathrm{2}} =\:\mathrm{4}\alpha\:+\mathrm{1}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\alpha^{\:\mathrm{3}} +\:\mathrm{17}\beta\:+\mathrm{5}\:=\:\alpha\:.\:\alpha^{\:\mathrm{2}} +\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\alpha\:\left(\:\mathrm{4}\alpha\:+\mathrm{1}\:\right)+\:\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\mathrm{4}\alpha^{\:\mathrm{2}} +\:\alpha\:+\:\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\mathrm{4}\:\left(\mathrm{4}\alpha\:+\mathrm{1}\:\right)+\alpha\:+\mathrm{17}\beta\:+\mathrm{5}=\mathrm{17}\left(\alpha+\beta\right)+\mathrm{9} \\ $$$$\:\:=\:\mathrm{17S}\:+\mathrm{9}=\:\mathrm{17}\:\left(\mathrm{4}\:\right)+\mathrm{9}=\mathrm{77} \\ $$$$ \\ $$

Question Number 162171 Answers: 0 Comments: 0

Question Number 162112 Answers: 1 Comments: 0

∫(( cos(x))/((1−cos(x))^2 ))dx=?

$$\int\frac{\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}{\left(\mathrm{1}−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\right)^{\mathrm{2}} }\boldsymbol{{dx}}=? \\ $$

Question Number 162099 Answers: 0 Comments: 0

PROVE THAT Ω= ∫_0 ^( 1) (( Li_( 2) (x ). ln( x ))/x) dx =^? (( −π^( 4) )/(90)) −−−−−−−−−− Ω= ∫_0 ^( 1) ln (x )Σ_(n=1) ^∞ (( x^( n−1) )/n^( 2) ) dx = Σ_(n=1) ^∞ (1/n^( 2) ) ∫_0 ^( 1) x^( n−1) . ln(x ) dx = Σ_(n=1) ^∞ (1/n^( 2) ) {[ (x^( n) /n) ln( x )]_0 ^( 1) −(1/n) ∫_0 ^( 1) x^( n−1) dx} = Σ_(n=1) ^∞ ((−1)/n^( 4) ) = − ζ (4 ) = ((−π^( 4) )/( 90)) ■ m.n −−− M . N −−−

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathscr{PROVE}\:\:\:\mathscr{THAT}\:\: \\ $$$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{Li}_{\:\mathrm{2}} \:\left({x}\:\right).\:\mathrm{ln}\left(\:{x}\:\right)}{{x}}\:{dx}\:\overset{?} {=}\:\frac{\:−\pi^{\:\mathrm{4}} }{\mathrm{90}} \\ $$$$\:\:\:\:\:−−−−−−−−−− \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\:\left({x}\:\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{x}^{\:{n}−\mathrm{1}} }{{n}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\:{n}−\mathrm{1}} .\:\mathrm{ln}\left({x}\:\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\:\mathrm{2}} }\:\left\{\left[\:\frac{{x}^{\:{n}} }{{n}}\:\mathrm{ln}\left(\:{x}\:\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:{n}−\mathrm{1}} {dx}\right\} \\ $$$$\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{1}}{{n}^{\:\mathrm{4}} }\:=\:−\:\zeta\:\left(\mathrm{4}\:\right)\:=\:\frac{−\pi^{\:\mathrm{4}} }{\:\mathrm{90}}\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−\:\mathscr{M}\:.\:\mathscr{N}\:\:−−−\: \\ $$$$ \\ $$

Question Number 162117 Answers: 2 Comments: 1

Question Number 162066 Answers: 0 Comments: 0

Σ_(n=1) ^∞ (((−1)^n H_n )/n^2 )=???

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }=??? \\ $$

Question Number 162055 Answers: 1 Comments: 0

∫e^(2x) (√((1 −e^(2x) )))dx

$$\int{e}^{\mathrm{2x}} \sqrt{\left(\mathrm{1}\:−{e}^{\mathrm{2}{x}} \right)}{dx} \\ $$

Question Number 162054 Answers: 1 Comments: 0

Question Number 162073 Answers: 3 Comments: 0

prove that Ω =∫_(−∞) ^( +∞) (( cos (x))/((2+ 2x +x^( 2) )^( 2) )) dx = (π/e) cos(1)

$$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\Omega\:=\int_{−\infty} ^{\:+\infty} \frac{\:{cos}\:\left({x}\right)}{\left(\mathrm{2}+\:\mathrm{2}{x}\:+{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{{e}}\:{cos}\left(\mathrm{1}\right) \\ $$

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