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∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)

Question Number 85670 Answers: 0 Comments: 3

Question Number 85669 Answers: 1 Comments: 4

∫ ((√(1+x))/(√(1−x))) dx

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

Question Number 85668 Answers: 0 Comments: 2

z = 2 + i find arg(z)

$$\mathrm{z}\:=\:\mathrm{2}\:+\:\mathrm{i}\: \\ $$$$\mathrm{find}\:\mathrm{arg}\left(\mathrm{z}\right) \\ $$

Question Number 85667 Answers: 1 Comments: 0

∫ (dx/(√(1−sin 2x)))

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

Question Number 85666 Answers: 0 Comments: 0

Question Number 85664 Answers: 0 Comments: 2

Show that a group order 100 is not simple

$$\boldsymbol{{S}\mathrm{how}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{order}}\:\mathrm{100}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{simple}} \\ $$

Question Number 85656 Answers: 0 Comments: 1

Question Number 85655 Answers: 0 Comments: 0

Question Number 85653 Answers: 0 Comments: 1

if f≥0 and (d/dx)(f(x))^2 =(f′(x))^2 and f(0)=1 find f(x) if f(x)=((4x^3 )/(x^2 +1)) find f^( −1) (x) and (f^( −1) )^′ (2)

$${if}\:\:{f}\geqslant\mathrm{0}\:\:{and}\:\:\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)^{\mathrm{2}} =\left({f}'\left({x}\right)\right)^{\mathrm{2}} \:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${find}\:{f}\left({x}\right)\:\:\: \\ $$$$ \\ $$$${if}\:{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:\:{find}\:{f}^{\:−\mathrm{1}} \left({x}\right)\:\:\:{and}\:\left({f}^{\:−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right) \\ $$

Question Number 85649 Answers: 1 Comments: 1

Proove that : ((√(2−(√3)))/2) = (((√6)−(√2))/4)

$$\mathrm{Proove}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Question Number 85648 Answers: 2 Comments: 0

∫(((x^3 −4))/((x+1)))dx

$$\int\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{4}\right)}{\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$

Question Number 85646 Answers: 0 Comments: 0

show that ∫(1/([x(x−1)(x−2)(x−3)...(x−m)]^2 ))dx= =(1/((m!)^2 ))Σ_(n=0) ^m ( ((m),(n) )^2 /(n−x))+(2/((m!)^2 ))ln∣Π_(n=0) ^m (x−n)^( ((m),(n) )^2 (H_(m−n) −H_n )) ∣+c

$${show}\:{that} \\ $$$$\int\frac{\mathrm{1}}{\left[{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)...\left({x}−{m}\right)\right]^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\left({m}!\right)^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} }{{n}−{x}}+\frac{\mathrm{2}}{\left({m}!\right)^{\mathrm{2}} }{ln}\mid\underset{{n}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}−{n}\right)^{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \left({H}_{{m}−{n}} −{H}_{{n}} \right)} \mid+{c} \\ $$

Question Number 85641 Answers: 0 Comments: 2

calculate A_λ =∫_3 ^∞ (dx/((x+λ)^3 (x−2)^4 )) (λ>0)

$${calculate}\:{A}_{\lambda} =\int_{\mathrm{3}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\lambda\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{4}} }\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$

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