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... advanced calculus... i : prove that :: ∫_0 ^( 1) ((ln(1+ln(1−x)))/(ln(1−x))) dx =^? Σ_(n=1) ^∞ ((Γ(n+1))/n^2 ) ii: prove that :: Ω =∫_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx =^? ((5π^2 )/(48)) m.n.july 1970#

$$\:\:\:\:\:\:\:\:...\:\:{advanced}\:{calculus}... \\ $$$$ \\ $$$${i}\::\:\:{prove}\:\:{that}\::: \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{ln}\left(\mathrm{1}−{x}\right)\right)}{{ln}\left(\mathrm{1}−{x}\right)}\:{dx}\:\overset{?} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${ii}:\: \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:\overset{?} {=}\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.{july}\:\mathrm{1970}# \\ $$$$\:\:\: \\ $$

Question Number 114044 Answers: 1 Comments: 0

old and unanswered... Mr Mathdave??? ∫x^2 ln(1−x)ln(1+x)dx=?

$${old}\:{and}\:{unanswered}...\:{Mr}\:{Mathdave}??? \\ $$$$\int{x}^{\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right){dx}=? \\ $$

Question Number 113929 Answers: 2 Comments: 4

∫(x+1)^2 (1−x)^5 dx

$$\int\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{5}} \mathrm{dx}\: \\ $$$$ \\ $$

Question Number 113910 Answers: 4 Comments: 0

(1)∫_0 ^π ((sin^4 x)/((1+cos x)^2 )) dx ? (2) lim_(x→∞) ((√(1−cos (((2π)/x))))/(1/x)) ?

$$\left(\mathrm{1}\right)\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:^{\mathrm{4}} {x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:{dx}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{{x}}\right)}}{\frac{\mathrm{1}}{{x}}}\:? \\ $$

Question Number 113907 Answers: 1 Comments: 0

∫ (√x) cos ((√x)) dx

$$\int\:\sqrt{{x}}\:\mathrm{cos}\:\left(\sqrt{{x}}\right)\:{dx} \\ $$

Question Number 113868 Answers: 0 Comments: 1

Find the area between the circle ρ=2acosθ and cardiode ρ=a(1+cosθ)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circle}\:\rho=\mathrm{2acos}\theta\:\mathrm{and}\: \\ $$$$\mathrm{cardiode}\:\rho=\mathrm{a}\left(\mathrm{1}+\mathrm{cos}\theta\right) \\ $$

Question Number 113867 Answers: 1 Comments: 0

∫_3 ^6 ((x+1)/(x^3 +x^2 −6x))dx

$$\int_{\mathrm{3}} ^{\mathrm{6}} \frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{6x}}\mathrm{dx} \\ $$

Question Number 113865 Answers: 0 Comments: 0

Consider the series I_n =∫_1 ^e x(lnx)^n dx and I_0 =∫_1 ^e xdx Which of the following is true ? a\ 0≤I_n ≤(e^2 /(n+2)) b\1≤I_n ≤(e^2 /(n+1)) c\I_n is negative

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{series}\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{x}\left(\mathrm{lnx}\right)^{\mathrm{n}} \mathrm{dx}\:\mathrm{and}\:\mathrm{I}_{\mathrm{0}} =\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{xdx} \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:? \\ $$$$\mathrm{a}\backslash\:\mathrm{0}\leqslant\mathrm{I}_{\mathrm{n}} \leqslant\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\:\:\:\:\mathrm{b}\backslash\mathrm{1}\leqslant\mathrm{I}_{\mathrm{n}} \leqslant\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{n}+\mathrm{1}}\:\:\mathrm{c}\backslash\mathrm{I}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{negative} \\ $$

Question Number 113821 Answers: 3 Comments: 1

∫_0 ^(π/2) ln(2−sinx)dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}−\mathrm{sinx}\right)\mathrm{dx} \\ $$

Question Number 113766 Answers: 1 Comments: 0

I=∫_0 ^∞ ((π/(1+π^2 x^2 ))−(1/(1+x^2 )))lnx dx put πx=tanA, x =tanB I=∫_0 ^(π/2) (ln(tanA)−lnπ)dA−∫_0 ^(π/2) ln(tanB)dB I=((−π)/2)lnπ

$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\pi}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){lnx}\:{dx} \\ $$$${put}\:\pi{x}={tanA},\:{x}\:={tanB} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({ln}\left({tanA}\right)−{ln}\pi\right){dA}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({tanB}\right){dB} \\ $$$${I}=\frac{−\pi}{\mathrm{2}}{ln}\pi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 113760 Answers: 1 Comments: 0

∫(√((x−1)/x^5 ))dx

$$\int\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }}\mathrm{dx} \\ $$

Question Number 113745 Answers: 1 Comments: 0