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Question Number 210208 Answers: 4 Comments: 0

∫_0 ^1 e^e^e^x e^e^x e^x dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{e}^{{e}^{{x}} } } \:{e}^{{e}^{{x}} } \:{e}^{{x}} {dx} \\ $$$$ \\ $$

Question Number 210206 Answers: 0 Comments: 0

Ω=∫_(1/e) ^e (dx/((1+x^2 )(1+xlog^7 x)))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\mathrm{log}\:^{\mathrm{7}} {x}\right)} \\ $$$$ \\ $$

Question Number 210229 Answers: 3 Comments: 0

Question Number 210228 Answers: 0 Comments: 0

Question Number 210227 Answers: 0 Comments: 0

Question Number 210369 Answers: 0 Comments: 0

Question Number 210194 Answers: 0 Comments: 1

Question Number 210181 Answers: 0 Comments: 3

Question Number 210180 Answers: 1 Comments: 2

Question Number 210172 Answers: 3 Comments: 0

Question Number 210171 Answers: 0 Comments: 0

Find: lim_(n→+∞) (n/((n!)^2 4^n )) Π_(k=1) ^n ((2k−1)^2 + 4) = ?

$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\mathrm{lim}}\:\:\frac{\mathrm{n}}{\left(\mathrm{n}!\right)^{\mathrm{2}} \:\mathrm{4}^{\boldsymbol{\mathrm{n}}} }\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\left(\left(\mathrm{2k}−\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{4}\right)\:=\:? \\ $$

Question Number 210157 Answers: 3 Comments: 0

Question Number 210156 Answers: 1 Comments: 0

Question Number 210155 Answers: 1 Comments: 0

Question Number 210142 Answers: 0 Comments: 1

Question Number 210133 Answers: 1 Comments: 0

calcul ∫_0 ^1 [nt^(n−1) (1−t)−t^n ]dt

$${calcul} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[{nt}^{{n}−\mathrm{1}} \left(\mathrm{1}−{t}\right)−{t}^{{n}} \right]{dt} \\ $$

Question Number 210127 Answers: 1 Comments: 0

Question Number 210126 Answers: 1 Comments: 0

Question Number 210124 Answers: 0 Comments: 0

Question Number 210120 Answers: 2 Comments: 0

Question Number 210112 Answers: 2 Comments: 0

show that ∫_0 ^∞ (x^n /((x+1)(ax+b)))dx=((((b/a))^n −1)/(b−a))𝛑csc(𝛑n) a>0,b>0,∣n∣<1 guys kill this let me see

$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{b}}\right)}\boldsymbol{\mathrm{dx}}=\frac{\left(\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}\right)^{\boldsymbol{\mathrm{n}}} −\mathrm{1}}{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}\boldsymbol{\pi\mathrm{csc}}\left(\boldsymbol{\pi\mathrm{n}}\right)\:\boldsymbol{\mathrm{a}}>\mathrm{0},\boldsymbol{\mathrm{b}}>\mathrm{0},\mid\boldsymbol{\mathrm{n}}\mid<\mathrm{1} \\ $$$$\boldsymbol{\mathrm{guys}}\:\boldsymbol{\mathrm{kill}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{see}} \\ $$

Question Number 210098 Answers: 5 Comments: 4

show that ∫_0 ^1 ((lnx)/(x^2 −1))dx=(𝛑^2 /8)

$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}} \\ $$

Question Number 210095 Answers: 1 Comments: 1

Question Number 210091 Answers: 1 Comments: 2

find Σ_(n=1) ^∞ tan^(−1) ((1/(2n^2 )))=?

$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)=? \\ $$

Question Number 210087 Answers: 0 Comments: 0

Question Number 210085 Answers: 1 Comments: 0

(1/((1/(2003))+(1/(2004))+(1/(2005))+(1/(2006))+(1/(2007))+(1/(2008))+(1/(2009)))) = ? Help me

$$ \\ $$$$\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2003}}+\frac{\mathrm{1}}{\mathrm{2004}}+\frac{\mathrm{1}}{\mathrm{2005}}+\frac{\mathrm{1}}{\mathrm{2006}}+\frac{\mathrm{1}}{\mathrm{2007}}+\frac{\mathrm{1}}{\mathrm{2008}}+\frac{\mathrm{1}}{\mathrm{2009}}}\:=\:? \\ $$$$\:\:\:\mathscr{H}{elp}\:{me} \\ $$$$ \\ $$

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