Question Number 130322 by mr W last updated on 24/Jan/21 $${to}\:{TinkuTara} \\ $$$${dear}\:{sir}: \\ $$$${can}\:{you}\:{please}\:{check}\:{following}\:{issue}: \\ $$$${what}\:{could}\:{be}\:{the}\:{reason}\:{that}\:{i}\:{can}'{t} \\ $$$${access}\:{to}\:{my}\:{old}\:{bookmarked}\:{posts} \\ $$$${before}\:{a}\:{special}\:{date}? \\ $$ Commented by…
Question Number 130317 by stelor last updated on 24/Jan/21 Answered by mathmax by abdo last updated on 24/Jan/21 $$\mathrm{I}=\int_{−\infty} ^{+\infty} \:\mathrm{x}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}}}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}=\int_{−\infty}…
Question Number 130315 by stelor last updated on 24/Jan/21 Answered by Dwaipayan Shikari last updated on 24/Jan/21 $$\int_{−\infty} ^{\infty} {xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\:\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}={u}\Rightarrow{x}=\frac{{du}}{{dx}} \\ $$$$=\int_{−\infty}…
Question Number 130303 by stelor last updated on 24/Jan/21 $$\mathrm{hello}…\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{limited}\:\mathrm{development}\:\mathrm{of}…{f}…\:{in}\:\mathrm{0}\:{at}\:\mathrm{3}^{{rd}} \:{order}. \\ $$$$\:\:\:\:{f}\left({x}\right)={ln}\left({sin}\left({x}\right)\right) \\ $$$$ \\ $$ Commented by stelor last updated on 24/Jan/21 $$\mathrm{developpement}\:\mathrm{limite}\:\mathrm{en}\:\frac{\Pi}{\mathrm{2}}\:\mathrm{a}\:\mathrm{l}'\mathrm{ordre}\:\mathrm{3}.…
Question Number 130287 by naka3546 last updated on 24/Jan/21 $$\int\:\frac{{x}}{{x}^{\mathrm{4}} +\mathrm{1}}\:{dx}\:\:=\:\:? \\ $$ Answered by mathmax by abdo last updated on 24/Jan/21 $$\mathrm{I}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow…
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Question Number 130255 by stelor last updated on 23/Jan/21 $$\mathrm{please}\:\:….\:\mathrm{I}\:\mathrm{am}\:\mathrm{confiouse}….\:\:\:\mathrm{1}\:\mathrm{and}\:\mathrm{2}. \\ $$$$\mathrm{1}.\:\:\:\:\:\:\:\int\mid{x}\mid\mathrm{d}{x}\:=\:?\:? \\ $$$$\mathrm{2}.\:\:\:\:\:\mathrm{with}\:\left(\mathrm{F}\left({x}\right)\right)^{'} \:=\mathrm{f}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\mid\mathrm{f}\left({x}\right)\mid{dx}\:=\:?? \\ $$$$ \\ $$ Answered by Olaf last updated on…
Question Number 130245 by mohammad17 last updated on 23/Jan/21 Commented by mohammad17 last updated on 23/Jan/21 $${please}\:{sir}\:{can}\:{help}\:{me}\:{now}\:{i}\:{want}\:{this}? \\ $$ Answered by PRITHWISH SEN 2 last…
Question Number 130226 by stelor last updated on 23/Jan/21 $$\mathrm{please}\:\mathrm{I}\:\mathrm{need}\:\mathrm{help}… \\ $$$$\int\left(\frac{\mathrm{e}^{{x}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}}\right){dx} \\ $$ Answered by Lordose last updated on 23/Jan/21 $$\Omega\:\overset{\mathrm{u}=\mathrm{e}^{\mathrm{x}} }…