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IntegrationQuestion and Answers: Page 134

Question Number 119821 Answers: 3 Comments: 0

∫_(−3) ^0 ((6x−6)/( (√(x^2 −2x+1)))) dx =?

$$\:\underset{−\mathrm{3}} {\overset{\mathrm{0}} {\int}}\:\frac{\mathrm{6}{x}−\mathrm{6}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}}\:{dx}\:=? \\ $$

Question Number 119762 Answers: 3 Comments: 0

calculate ∫_0 ^∞ ((x^4 dx)/((2x+1)^5 (3x+1)^8 ))

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{x}^{\mathrm{4}} \mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{8}} } \\ $$

Question Number 119752 Answers: 1 Comments: 0

find I_λ =∫_0 ^∞ ((ch(1+λcosx))/((x^2 +1)^2 ))dx (λ real >0)

$${find}\:{I}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left(\mathrm{1}+\lambda{cosx}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left(\lambda\:{real}\:>\mathrm{0}\right) \\ $$

Question Number 119750 Answers: 1 Comments: 0

Π_(k=1) ^(1019) [((2k)/(2k−1))]=?

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{1019}} {\prod}}\left[\frac{\mathrm{2k}}{\mathrm{2k}−\mathrm{1}}\right]=? \\ $$

Question Number 119696 Answers: 2 Comments: 0

For a<b then ∫_a ^b (x−a)(x−b) dx equal to _

$${For}\:{a}<{b}\:{then}\:\underset{{a}} {\overset{{b}} {\int}}\:\left({x}−{a}\right)\left({x}−{b}\right)\:{dx}\: \\ $$$${equal}\:{to}\:\_ \\ $$

Question Number 119681 Answers: 4 Comments: 0

∫_0 ^π (√((1+cos2x)/2)) dx ∫_0 ^∞ [ne^(−x) ]dx

$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\ $$

Question Number 119570 Answers: 3 Comments: 0

... ♣_♣ ^♣ nice calculus♣_♣ ^♣ ... prove that :: Ω=∫_0 ^( ∞) e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))=1 ...★ M.N.1970★...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}{nice}\:\:{calculus}\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}{x}} {ln}\left(\frac{\mathrm{1}+{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\bigstar\:\mathscr{M}.\mathscr{N}.\mathrm{1970}\bigstar... \\ $$

Question Number 119591 Answers: 0 Comments: 6

...nice calculus... prove that :: ∫_0 ^( (π/2)) (√(((2^x −1)sin^3 (x))/((2^x +1)(sin^3 (x)+cos^3 (x))))) dx<(π/8) ...m.n.1970...

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\left(\mathrm{2}^{{x}} −\mathrm{1}\right){sin}^{\mathrm{3}} \left({x}\right)}{\left(\mathrm{2}^{{x}} +\mathrm{1}\right)\left({sin}^{\mathrm{3}} \left({x}\right)+{cos}^{\mathrm{3}} \left({x}\right)\right)}}\:\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$ \\ $$

Question Number 119462 Answers: 2 Comments: 0

... advanced calculus... evaluate :: Ω=∫_0 ^( ∞) ((tan^(−1) (x))/(e^(2πx) −1))dx =? m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119446 Answers: 0 Comments: 0

calculste ∫_0 ^∞ ((ln(2+x^2 ))/(1+x^3 ))dx

$${calculste}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$

Question Number 119445 Answers: 0 Comments: 0

...nice calculus... prove that:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(n^3 (((2n)),(n) ))) =^(???) ζ(3) m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119442 Answers: 0 Comments: 0

... advanced calculus... prove that : Σ_(n=1) ^∞ (1/(n^2 (((2n)),(n) ))) =^(???) ((ζ(2))/3) solution:: Σ_(n=1) ^∞ (1/(n^2 ∗(((2n)!)/((n!)^2 ))))=Σ_(n=1) ^∞ ((n!∗n!)/(n^2 ∗(2n)!)) =Σ_(n=1) ^∞ ((Γ(n)Γ(n+1))/(nΓ(2n+1)))=Σ_(n=1) ^∞ ((β(n,n+1))/n) =Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n−1) (1−x)^n =∫_0 ^( 1) (1/x)Σ(((x−x^2 )^n )/n)dx =−∫_0 ^( 1) ((ln(1−x+x^2 ))/x)dx =−∫_0 ^( 1) ((ln(1+x^3 )−ln(1+x))/x)dx =∫_0 ^( 1) ((ln(1+x))/x)dx −∫_0 ^( 1) ((ln(1+x^3 ))/x)dx =−li_2 (−1) −∫_0 ^( 1) ((Σ_(n=1) (((−1)^(n+1) x^(3n) )/n))/x) dx =(π^2 /(12))+Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^( 1) x^(3n−1) dx =(π^2 /(12)) +(1/3)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(12))−(π^2 /(36)) =(π^2 /(18)) =((ζ(2))/3) ✓✓ m.n.july.1970..

$$\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${solution}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \ast\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!\ast{n}!}{{n}^{\mathrm{2}} \ast\left(\mathrm{2}{n}\right)!}\: \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\Gamma\left({n}\right)\Gamma\left({n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left({n},{n}+\mathrm{1}\right)}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}}\Sigma\frac{\left({x}−{x}^{\mathrm{2}} \right)^{{n}} }{{n}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}{dx} \\ $$$$\:=−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\underset{{n}=\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}} }{{n}}}{{x}}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{1}} {dx}\:\:\:\:\: \\ $$$$ \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}.. \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 119425 Answers: 1 Comments: 0

decompose F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) and calculate ∫_(√2) ^(+∞) F(x)dx

$$\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{2x}−\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$$\mathrm{and}\:\mathrm{calculate}\:\int_{\sqrt{\mathrm{2}}} ^{+\infty} \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 119335 Answers: 2 Comments: 0

Express f(x) = (1/((x−1)^2 (x^2 +1))) into partial fractions. hence evaluate I = ∫_0 ^4 f(x) dx

$$\:\mathrm{Express}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:\:\mathrm{into}\:\mathrm{partial}\:\mathrm{fractions}. \\ $$$$\mathrm{hence}\:\mathrm{evaluate}\:{I}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right)\:{dx} \\ $$

Question Number 119323 Answers: 1 Comments: 4

Question Number 119306 Answers: 3 Comments: 0

Question Number 119282 Answers: 1 Comments: 0

... nice calculus... evaluate:: I:= ∫_0 ^( 1) li_2 (1−x^2 )dx=?? .m.n.1970.

$$\:\:\:\:\:\:\:\:...\:\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{evaluate}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}=?? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 119318 Answers: 0 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^4 +x^2 +2))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{lnx}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{2}}{dx} \\ $$

Question Number 119317 Answers: 0 Comments: 0

calculate ∫_0 ^(2π) (dθ/((x^2 −2cosθ x+1)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{d}\theta}{\left({x}^{\mathrm{2}} −\mathrm{2}{cos}\theta\:{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 119262 Answers: 3 Comments: 0

∫ (x^2 /( (√((4−x^2 )^5 )))) dx

$$\:\int\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{4}−{x}^{\mathrm{2}} \right)^{\mathrm{5}} }}\:{dx}\: \\ $$

Question Number 119256 Answers: 0 Comments: 0

Determine ∫_(−(π/4)) ^( (π/4)) (cost+(√(1+t^2 sin^3 tcos^3 t))dt=?

$$\: \\ $$$$\:\:\:\:\:\:{Determine}\:\:\int_{−\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{4}}} \left({cost}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} {sin}^{\mathrm{3}} {tcos}^{\mathrm{3}} {t}}{dt}=?\right. \\ $$

Question Number 119151 Answers: 0 Comments: 0

Question Number 119070 Answers: 2 Comments: 0

∫ ((x^2 −x+6)/(x^3 +3x)) dx ∫ ((5x^2 +3x−2)/(x^3 +2x^2 )) dx

$$\int\:\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{3}{x}}\:{dx}\: \\ $$$$\int\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }\:{dx}\: \\ $$

Question Number 119038 Answers: 2 Comments: 0

∫_(−π/4) ^(+π/4) ((√(1+tan x))/( (√(1−tan x))))dx

$$\underset{−\pi/\mathrm{4}} {\overset{+\pi/\mathrm{4}} {\int}}\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{dx} \\ $$

Question Number 119021 Answers: 1 Comments: 0

Question Number 119006 Answers: 2 Comments: 0

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