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Question Number 98679 Answers: 1 Comments: 2

prove that ∫_0 ^∞ ((3+2(√x))/(x^2 +2x+5))dx=4.13049

$${prove}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}+\mathrm{2}\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\mathrm{4}.\mathrm{13049}\: \\ $$

Question Number 98678 Answers: 2 Comments: 9

∫_0 ^∞ e^(−ax) ((sin mx)/x) dx = tan^(−1) ((m/a)), a>0

$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ax}}} \frac{\mathrm{sin}\:\boldsymbol{\mathrm{mx}}}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{a}}}\right),\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98677 Answers: 1 Comments: 0

prove ∫_0 ^a ((ln(1+ax))/(1+x^2 ))dx=(1/2)ln(1+a^2 )tan^(−1) a, a>0

$$\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{ax}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98675 Answers: 0 Comments: 10

Question Number 98673 Answers: 2 Comments: 0

find a_n in terms of n (I can′t find it...) a_1 =1; a_2 =4 a_3 =a_2 ×4×((2^2 −1)/2^2 ) a_4 =a_3 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 ) a_5 =a_4 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 )×((4^2 −1)/4^2 ) ... n≥2: a_(n+1) =4a_n Π_(k=2) ^n ((k^2 −1)/k^2 )

$$\mathrm{find}\:{a}_{{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\left(\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}...\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$... \\ $$$${n}\geqslant\mathrm{2}:\:{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 98672 Answers: 1 Comments: 0

∫_0 ^4 ∫_0 ^(x/4) e^x^2 dx dy

$$\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\mathrm{0}} ^{\frac{{x}}{\mathrm{4}}} {e}^{{x}^{\mathrm{2}} } \:{dx}\:{dy} \\ $$

Question Number 98661 Answers: 0 Comments: 1

using cayley − hamilton theorem what is the inverse of matrix A= [((0 1 −1)),((1 2 2)),((0 1 −1)) ]

$$\mathrm{using}\:\mathrm{cayley}\:−\:\mathrm{hamilton} \\ $$$$\mathrm{theorem}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\mathrm{matrix}\:\mathrm{A}=\:\begin{bmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{bmatrix}\: \\ $$

Question Number 98796 Answers: 2 Comments: 0

Question Number 98657 Answers: 2 Comments: 0

solve y^(′′) −3y^′ +2y =((sinx)/x)

$$\mathrm{solve}\:\:\mathrm{y}^{''} \:−\mathrm{3y}^{'} \:\:+\mathrm{2y}\:=\frac{\mathrm{sinx}}{\mathrm{x}} \\ $$

Question Number 98656 Answers: 2 Comments: 0

solve xy^(′′) +(2+x^2 )y^′ =xe^(−x^2 )

$$\mathrm{solve}\:\:\:\mathrm{xy}^{''} \:+\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}^{'} \:\:=\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 98653 Answers: 0 Comments: 0

Question Number 98640 Answers: 0 Comments: 1

Question Number 98638 Answers: 0 Comments: 2

Le plan complexe est rapporte^ a^ un repe^ re orthornorme directe (0,e_1 ^→ ,e_2 ^→ ). On note A et B les points d′affixes respectives i, et 2i. Soit f, l′application du plan prive^ de A dans lui-me^ me qui a^ tout point M d′affixe z distincte i associe le point M d′affixe z′ definie par: z′=((2z−i)/(iz+1)) 1\ Soit z≠i a\ On pose z−i=re^(iθ) . Interpreter ge^ ometriquement r et θ a^ l′aide des points A et M.

$$\mathrm{Le}\:\mathrm{plan}\:\mathrm{complexe}\:\mathrm{est}\:\mathrm{rapport}\acute {\mathrm{e}}\:\grave {\mathrm{a}}\:\mathrm{un}\:\mathrm{rep}\grave {\mathrm{e}re} \\ $$$$\mathrm{orthornorme}\:\mathrm{directe}\:\left(\mathrm{0},\overset{\rightarrow} {\mathrm{e}}_{\mathrm{1}} ,\overset{\rightarrow} {\mathrm{e}}_{\mathrm{2}} \right).\:\mathcal{O}\mathrm{n}\:\mathrm{note}\:\mathrm{A}\:\mathrm{et}\:\mathrm{B}\:\mathrm{les} \\ $$$$\mathrm{points}\:\mathrm{d}'\mathrm{affixes}\:\mathrm{respectives}\:\boldsymbol{\mathrm{i}},\:\mathrm{et}\:\mathrm{2}\boldsymbol{\mathrm{i}}.\:\mathrm{Soit}\:\mathrm{f},\:\mathrm{l}'\mathrm{application} \\ $$$$\mathrm{du}\:\mathrm{plan}\:\mathrm{priv}\acute {\mathrm{e}}\:\mathrm{de}\:\mathrm{A}\:\mathrm{dans}\:\mathrm{lui}-\mathrm{m}\hat {\mathrm{e}me}\:\mathrm{qui}\:\grave {\mathrm{a}}\:\mathrm{tout}\:\mathrm{point} \\ $$$$\mathrm{M}\:\mathrm{d}'\mathrm{affixe}\:\mathrm{z}\:\mathrm{distincte}\:\boldsymbol{\mathrm{i}}\:\mathrm{associe}\:\mathrm{le}\:\mathrm{point}\:\mathrm{M}\:\mathrm{d}'\mathrm{affixe} \\ $$$$\boldsymbol{\mathrm{z}}'\:\mathrm{definie}\:\mathrm{par}:\:\mathrm{z}'=\frac{\mathrm{2z}−\mathrm{i}}{\mathrm{iz}+\mathrm{1}} \\ $$$$\mathrm{1}\backslash\:\mathrm{Soit}\:\mathrm{z}\neq\mathrm{i} \\ $$$$\mathrm{a}\backslash\:\mathrm{On}\:\mathrm{pose}\:\mathrm{z}−\mathrm{i}=\mathrm{re}^{\mathrm{i}\theta} .\:\mathcal{I}\mathrm{nterpreter}\:\mathrm{g}\acute {\mathrm{e}ometriquement}\:\mathrm{r}\:\mathrm{et}\:\theta \\ $$$$\grave {\mathrm{a}}\:\mathrm{l}'\mathrm{aide}\:\mathrm{des}\:\mathrm{points}\:\mathrm{A}\:\mathrm{et}\:\mathrm{M}. \\ $$

Question Number 98623 Answers: 3 Comments: 0

evaluate ∫_(2/(√3)) ^2 (1/(x^2 (√(4+x^2 ))))dx using the substitution x=2tanθ

$${evaluate}\: \\ $$$$\int_{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}} ^{\mathrm{2}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}+{x}^{\mathrm{2}} }}{dx}\:{using}\:{the}\:{substitution}\:{x}=\mathrm{2tan}\theta \\ $$$$ \\ $$

Question Number 98621 Answers: 2 Comments: 1