Question Number 108071 by mathdave last updated on 14/Aug/20 Answered by mathmax by abdo last updated on 14/Aug/20 $$\mathrm{sir}\:\mathrm{consider}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{a}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{f}^{'}…
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Question Number 173603 by Gbenga last updated on 14/Jul/22 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{log}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108051 by ZiYangLee last updated on 14/Aug/20 Answered by Sarah85 last updated on 14/Aug/20 $$\sqrt{{x}}−\sqrt{{y}}\geqslant\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\sqrt{{y}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}−\mathrm{2}\sqrt{{xy}}+{y}\geqslant\mathrm{0} \\ $$$${x}+{y}\geqslant\mathrm{2}\sqrt{{xy}} \\…
Question Number 108043 by ZiYangLee last updated on 14/Aug/20 $$\mathrm{Given}\:\mathrm{f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}'\left(\mathrm{0}\right). \\ $$ Answered by hgrocks last updated on 14/Aug/20 $$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{x}+\mathrm{k}\right) \\…
Question Number 108034 by mathdave last updated on 14/Aug/20 Answered by mathmax by abdo last updated on 14/Aug/20 $$\mathrm{I}\:=\int_{\mathrm{49}} ^{\mathrm{196}} \:\:\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}}−\mathrm{2}\right)\sqrt{\mathrm{2}+\sqrt{\mathrm{x}}}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{tbe}\:\mathrm{cha7gement}\:\:\sqrt{\mathrm{x}}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\mathrm{7}} ^{\mathrm{14}} \:\frac{\mathrm{2t}\:\mathrm{dt}}{\left(\mathrm{t}−\mathrm{2}\right)\sqrt{\mathrm{2}+\mathrm{t}}}\:\mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{7}}…
Question Number 173566 by Khalmohmmad last updated on 13/Jul/22 Answered by MJS_new last updated on 13/Jul/22 $${x}=\mathrm{0}\vee{x}=−\infty \\ $$ Answered by mr W last updated…
Question Number 108016 by mathdave last updated on 13/Aug/20 Answered by mr W last updated on 13/Aug/20 $$\theta=\mathrm{cos}^{−\mathrm{1}} {x}\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\pi\:\Rightarrow\mathrm{0}\leqslant\mathrm{3}\theta\leqslant\mathrm{3}\pi \\ $$$$\mathrm{cos}\:\theta={x} \\ $$$$\phi=\mathrm{sin}^{−\mathrm{1}} \mathrm{2}{x}\:\Rightarrow−\frac{\pi}{\mathrm{2}}\leqslant\phi\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow−\pi\leqslant\mathrm{2}\phi\leqslant\pi \\…
Question Number 108017 by mathdave last updated on 13/Aug/20 Answered by hgrocks last updated on 13/Aug/20 $$\mathrm{2}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{y}^{\mathrm{2}} \Rightarrow\:\sqrt{\mathrm{x}}\:=\:\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{x}}}}\:\mathrm{dx}\:=\:\mathrm{dy} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{x}}}}\:\mathrm{dx}\:=\:\mathrm{4}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{2}\right) \\…
Question Number 42479 by mondodotto@gmail.com last updated on 26/Aug/18 $$\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{digit}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{seven}}\:\boldsymbol{\mathrm{times}}\: \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{digits}}.\boldsymbol{\mathrm{if}}\:\mathrm{27}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{substracted}}\: \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{digits}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{interchanged}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{number}} \\ $$$$ \\ $$ Answered by $@ty@m last updated…