Question Number 219606 by Nicholas666 last updated on 29/Apr/25 $$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by…
Question Number 219600 by Nicholas666 last updated on 29/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \left[{x}\right]^{{x}} \:{dx} \\ $$$$ \\ $$ Answered by Nicholas666 last updated on…
Question Number 219602 by SdC355 last updated on 29/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {J}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)−\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {Y}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)=?? \\ $$ Commented by Nicholas666 last updated…
Question Number 219597 by SdC355 last updated on 29/Apr/25 $$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$ Answered by Nicholas666 last updated on 29/Apr/25 $$\frac{\mathrm{2}\pi}{{n}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}}…
Question Number 219589 by Nicholas666 last updated on 29/Apr/25 $${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right)…
Question Number 219591 by SdC355 last updated on 29/Apr/25 $$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\…
Question Number 219586 by Hery03 last updated on 29/Apr/25 $${Integrate}\:: \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:−\:{x}}\:+\:\mathrm{1}}{dx}. \\ $$ Answered by Ghisom last updated on 29/Apr/25 $$\int\frac{{x}^{\mathrm{2}} }{\:\mathrm{1}+\sqrt{{x}}\sqrt{{x}−\mathrm{1}}}{dx}=…
Question Number 219587 by Nicholas666 last updated on 29/Apr/25 Commented by Nicholas666 last updated on 29/Apr/25 https://www.quora.com/profile/Bekicot-5/math-math-If-math-a-b-0-math-and-math-a-b-frac-2-3-math-Then-math-frac-b-3-3a-2-frac-a?ch=10&oid=221057452&share=32d38051&srid=5Xg5SU&target_type=post Answered by Ghisom last updated on 29/Apr/25 $${a},\:{b}>\mathrm{0}\wedge{a}+{b}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\:\mathrm{0}<{a},\:{b}<\frac{\mathrm{2}}{\mathrm{3}}…
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Question Number 219549 by Spillover last updated on 28/Apr/25 Answered by Nicholas666 last updated on 29/Apr/25 $$\int\frac{\mathrm{1}+{sin}\:{x}}{\mathrm{1}−{sinx}}{dx}=\mathrm{2}{tany}−{x}+{c} \\ $$$$\int\frac{\left(\mathrm{1}+{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{1}−{sin}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}\:+{c} \\ $$$$\int\frac{\mathrm{1}+\mathrm{2}\:{sin}\:{x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}+{c}…