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Question Number 221838 Answers: 2 Comments: 0

∫_0 ^∞ (((arctan x)/x))^2 dx=?

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{arctan}\:{x}}{{x}}\right)^{\mathrm{2}} {dx}=? \\ $$

Question Number 221832 Answers: 0 Comments: 0

f′(x)=lim_(h→0) ((f(x+h)−f(x))/h) lim_(h→0) ((c−c)/h)=lim_(h→0) 0=0 lim_(h→0) (((x+h)^2 −x^2 )/h)=lim_(h→0) ((2xh+h^2 )/h)=lim_(h→0) (2x+h)=2x lim_(h→0) (((x+h)^n −x^n )/h)=lim_(h→0) ((Σ_(k=0) ^n ((n),(k) )x^(n−k) h^k −x^n )/h)=lim_(h→0) ((nx^(n−1) h+ ((n),(2) )x^(n−2) h^2 +…)/h)=lim_(h→0) (nx^(n−1) + ((n),(2) )x^(n−2) h+…) lim_(h→0) ((e^(x+h) −e^x )/h)=e^x lim_(h→0) (e^(h−1) /h)=e^x ∙1 lim_(h→0) ((e^h −1)/h)=lim_(k→0) (k/(ln(1+k)))=lim_(k→0) (1/((ln(1+k))/k))=lim_(k→0) (1/(ln((1+k)^(1/k) ))) =(1/(ln e))=(1/1)=1 lim_(h→0) ((sin(x+h)−sin x)/h)=lim_(h→0) ((sin x cos h+cos x sin h−sin x)/h)=lim_(h→0) (sin x((cos h−1)/h)+cos x((sin h)/h)) lim_(h→0) ((sin h)/h)=1 lim_(h→0) ((cos h−1)/h)=lim_(h→0) ((−2 sin^2 (h/2))/h)=lim_(h→0) −((sin^2 (h/2))/((h/2)^2 ))∙(((h/2)^2 )/h)=lim_(h→0) −(((sin(h/2))/(h/2)))^2 ∙(h/4)=−1∙0=0 lim_(h→0) (sin x∙0+cos x∙1)=cos x ∫_a ^b f(x)dx=lim_(n→∞) Σ_(i=1) ^n f(a+i((b−a)/n))((b−a)/n) ∫_a ^b x^m dx=lim_(n→∞) Σ_(k=1) ^n (k(b/n))^m (b/n)=b^(m+1) lim_(n→∞) (1/n)Σ_(k=1) ^n ((k/n))^m lim_(n→∞) Σ_(k=1) ^n ((k/n))^m (b/n)=lim_(n→∞) (((n^(m+1) /(m+1))+O(n^m ))/n^(m+1) )=lim_(n→∞) ((1/(m+1))+O((1/n)))=(1/(m+1)) ∫_a ^b x^m dx=b^(m+1) ∙(1/(m+1))=(b^(m+1) /(m+1)) ∫x^m dx=(x^(m+1) /(m+1))+C G(x)=∫_0 ^x f(t)dt G′(x)=lim_(h→0) ((G(x+h)−G(x))/h)=lim_(h→0) ((∫_x ^(x+h) f(t)dt)/h)=lim_(h→0) ((f(c_n )h)/h)=lim_(h→0) f(c_n )=f(x) ,c_h ∈[x,x+h] F(b)−F(a)=[∫_a ^b f(t)dt+C]−[∫_a ^a f(t)dt+C]=∫_a ^b f(t)dt−0=∫_a ^b f(t)dt

$${f}'\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{c}−{c}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}0}=\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{xh}+{h}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}{x}+{h}\right)=\mathrm{2}{x} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{{n}} −{x}^{{n}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}−{k}} {h}^{{k}} −{x}^{{n}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{nx}^{{n}−\mathrm{1}} {h}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}{x}^{{n}−\mathrm{2}} {h}^{\mathrm{2}} +\ldots}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({nx}^{{n}−\mathrm{1}} +\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}{x}^{{n}−\mathrm{2}} {h}+\ldots\right) \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+{h}} −{e}^{{x}} }{{h}}={e}^{{x}} \underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{h}−\mathrm{1}} }{{h}}={e}^{{x}} \centerdot\mathrm{1} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{h}} −\mathrm{1}}{{h}}=\underset{{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{k}}{\mathrm{ln}\left(\mathrm{1}+{k}\right)}=\underset{{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\frac{\mathrm{ln}\left(\mathrm{1}+{k}\right)}{{k}}}=\underset{{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{ln}\left(\left(\mathrm{1}+{k}\right)^{\mathrm{1}/{k}} \right)}\:=\frac{\mathrm{1}}{\mathrm{ln}\:{e}}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}+{h}\right)−\mathrm{sin}\:{x}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:{h}+\mathrm{cos}\:{x}\:\mathrm{sin}\:{h}−\mathrm{sin}\:{x}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:{x}\frac{\mathrm{cos}\:{h}−\mathrm{1}}{{h}}+\mathrm{cos}\:{x}\frac{\mathrm{sin}\:{h}}{{h}}\right) \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{h}}{{h}}=\mathrm{1} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{h}−\mathrm{1}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \left({h}/\mathrm{2}\right)}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{sin}^{\mathrm{2}} \left({h}/\mathrm{2}\right)}{\left({h}/\mathrm{2}\right)^{\mathrm{2}} }\centerdot\frac{\left({h}/\mathrm{2}\right)^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left(\frac{\mathrm{sin}\left({h}/\mathrm{2}\right)}{{h}/\mathrm{2}}\right)^{\mathrm{2}} \centerdot\frac{{h}}{\mathrm{4}}=−\mathrm{1}\centerdot\mathrm{0}=\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:{x}\centerdot\mathrm{0}+\mathrm{cos}\:{x}\centerdot\mathrm{1}\right)=\mathrm{cos}\:{x} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({a}+{i}\frac{{b}−{a}}{{n}}\right)\frac{{b}−{a}}{{n}} \\ $$$$\int_{{a}} ^{{b}} {x}^{{m}} {dx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}\frac{{b}}{{n}}\right)^{{m}} \frac{{b}}{{n}}={b}^{{m}+\mathrm{1}} \underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}}\right)^{{m}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}}\right)^{{m}} \frac{{b}}{{n}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{n}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}+{O}\left({n}^{{m}} \right)}{{n}^{{m}+\mathrm{1}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{m}+\mathrm{1}}+{O}\left(\frac{\mathrm{1}}{{n}}\right)\right)=\frac{\mathrm{1}}{{m}+\mathrm{1}} \\ $$$$\int_{{a}} ^{{b}} {x}^{{m}} {dx}={b}^{{m}+\mathrm{1}} \centerdot\frac{\mathrm{1}}{{m}+\mathrm{1}}=\frac{{b}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}} \\ $$$$\int{x}^{{m}} {dx}=\frac{{x}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}+{C} \\ $$$${G}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt} \\ $$$${G}'\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{G}\left({x}+{h}\right)−{G}\left({x}\right)}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{{x}} ^{{x}+{h}} {f}\left({t}\right){dt}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({c}_{{n}} \right){h}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({c}_{{n}} \right)={f}\left({x}\right)\:,{c}_{{h}} \in\left[{x},{x}+{h}\right] \\ $$$${F}\left({b}\right)−{F}\left({a}\right)=\left[\int_{{a}} ^{{b}} {f}\left({t}\right){dt}+{C}\right]−\left[\int_{{a}} ^{{a}} {f}\left({t}\right){dt}+{C}\right]=\int_{{a}} ^{{b}} {f}\left({t}\right){dt}−\mathrm{0}=\int_{{a}} ^{{b}} {f}\left({t}\right){dt} \\ $$

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

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