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Question Number 221831 Answers: 1 Comments: 0

∫ ((√x)/( (√(1+x^2 )) + (√(1−x^2 )))) dx

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

Question Number 221830 Answers: 0 Comments: 0

∫_0 ^1 ((√x)/( (√(1+x^2 )) + (√(1−x^2 )))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx} \\ $$$$ \\ $$

Question Number 221829 Answers: 4 Comments: 0

(√(70.71.72.73+1))

$$\sqrt{\mathrm{70}.\mathrm{71}.\mathrm{72}.\mathrm{73}+\mathrm{1}} \\ $$

Question Number 221815 Answers: 1 Comments: 0

Question Number 221814 Answers: 1 Comments: 0

Question Number 221788 Answers: 6 Comments: 0

Question Number 221787 Answers: 1 Comments: 0

(√((1−4x(√(1−4x^2 )))/2)) = 1−8x^2 x=?

$$\:\:\sqrt{\frac{\mathrm{1}−\mathrm{4x}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}{\mathrm{2}}}\:=\:\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=?\: \\ $$

Question Number 221786 Answers: 0 Comments: 0

Express ∫_0 ^( 1) z^z^z^⋰ dz closed form... z⇈^∞ =z^z^z^⋰

$$\:\mathrm{Express}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{z}^{{z}^{{z}^{\iddots} } } \:\mathrm{d}{z}\:\:\mathrm{closed}\:\mathrm{form}... \\ $$$${z}\upuparrows^{\infty} ={z}^{{z}^{{z}^{\iddots} } } \\ $$

Question Number 221782 Answers: 0 Comments: 0

(((81))^(1/((27))^(1/3^a ) ) )^((√4)) where a=4^0^4^3 and b=Σ_(n=1) ^6 n

$$\left(\sqrt[{\sqrt[{\mathrm{3}^{{a}} }]{\mathrm{27}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$$${where}\:{a}=\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } {and}\:{b}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{n} \\ $$

Question Number 221778 Answers: 0 Comments: 0

For ∀n∈N^∗ ,n≥3 Prove: ∫_0 ^1 (Σ_(2<p≤2n) e^(2πip+α) )^2 e^(−4πinα) dα>0,p is a prime number

$$\mathrm{For}\:\forall{n}\in\boldsymbol{{N}}^{\ast} ,{n}\geq\mathrm{3}\:\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{\mathrm{2}<{p}\leq\mathrm{2}{n}} {\sum}{e}^{\mathrm{2}\pi{ip}+\alpha} \right)^{\mathrm{2}} {e}^{−\mathrm{4}\pi{in}\alpha} {d}\alpha>\mathrm{0},{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number} \\ $$

Question Number 221770 Answers: 0 Comments: 0

Prove:∫_0 ^1 ((arcsinx)/(1+x^4 ))dx=(((√2)π^2 )/(16))−(((√2)π)/8)ln((√2)−1)+2Σ_(n=0) ^∞ (((−)^n )/((2n+1)))∣z_0 ∣^(2n+1) sin((π/4)−(2n+1)β)

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arcsin}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\frac{\sqrt{\mathrm{2}}\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{8}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\mid{z}_{\mathrm{0}} \mid^{\mathrm{2}{n}+\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\left(\mathrm{2}{n}+\mathrm{1}\right)\beta\right) \\ $$

Question Number 221769 Answers: 0 Comments: 0

∫_0 ^π (a/(a−cos^(2n) x))dx=?,a>1 ∫_0 ^π (2/(2−cos^4 x))dx=? lim_(m→∞) ∫_0 ^π ((cos^(2n) (2mx))/(a−cos^(2n) x))dx=?,a>1,n∈N^+

$$\int_{\mathrm{0}} ^{\pi} \frac{{a}}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{cos}^{\mathrm{4}} {x}}{dx}=? \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}^{\mathrm{2}{n}} \left(\mathrm{2}{mx}\right)}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1},{n}\in\mathbb{N}^{+} \\ $$

Question Number 221760 Answers: 1 Comments: 0

Question Number 221754 Answers: 1 Comments: 0

((81))^(1/((64))^(1/((27))^(1/( 3^4^0^4^3 )) ) ) )^((√4))

$$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$

Question Number 221733 Answers: 0 Comments: 19

Question Number 221721 Answers: 2 Comments: 0

Question Number 221707 Answers: 1 Comments: 13

Question Number 221697 Answers: 2 Comments: 0

Is (√i) an imaginary number (i=(√(−1))) answer with logic

$${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$

Question Number 221686 Answers: 3 Comments: 2

Question Number 221668 Answers: 0 Comments: 0

Question Number 221663 Answers: 0 Comments: 3

Prove:∫_0 ^1 Π_(k=1) ^∞ (1−x^k )dx=((4π(√3))/( (√(23))))∙((sinh(((√(23))π)/6))/(2 cosh(((√(23))π)/3)−1))

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{x}^{{k}} \right){dx}=\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{23}}}\centerdot\frac{\mathrm{sinh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{6}}}{\mathrm{2}\:\mathrm{cosh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{3}}−\mathrm{1}} \\ $$

Question Number 221661 Answers: 1 Comments: 1

Question Number 221647 Answers: 0 Comments: 1

solve for x. x^1 + x^2 + x^3 = 4096

$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$

Question Number 221638 Answers: 1 Comments: 0

Question Number 221637 Answers: 1 Comments: 0

Question Number 221626 Answers: 3 Comments: 0

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