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Author: Tinku Tara

1-1-x-2-lnx-dx-

Question Number 148285 by Khalmohmmad last updated on 26/Jul/21 $$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \mathrm{ln}{x}}{dx}=? \\ $$ Commented by ahmedNasser last updated on 26/Jul/21 $${let}\:{u}\:=\:{lnx}\:\:\:\:\therefore\:{du}\:=\:\frac{\mathrm{1}}{{x}}\:{dx}\:\:\:\:{x}\:=\:{e}^{{u}} \\ $$$$\int_{\mathrm{0}}…

Question-148284

Question Number 148284 by 0731619 last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\mathrm{6}!!\:=\:\mathrm{6}×\mathrm{4}×\mathrm{2}\:=\:\mathrm{48} \\ $$$$\mathrm{8}!!!!!\:=\:\mathrm{8}×\mathrm{3}\:=\:\mathrm{24} \\ $$$$\mathrm{9}!!!!\:=\:\mathrm{9}×\mathrm{5}×\mathrm{1}\:=\:\mathrm{45} \\ $$$$!\mathrm{4}\:=\:\mathrm{4}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}\right)\:=\:\mathrm{9} \\ $$$$!\mathrm{5}\:=\:\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}−+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}\right)\:=\:\mathrm{44}…

Spin-only-magnetic-moment-of-25-Mn-x-is-15-B-M-Then-the-value-of-x-is-i-did-following-1s-2-2s-2-2p-6-3s-2-3p-6-4s-2-3d-5-so-to-get-3-unpaired-electron-we-need-to-2-electron-so-x-2-book-

Question Number 17209 by sushmitak last updated on 02/Jul/17 $$\mathrm{Spin}\:\mathrm{only}\:\mathrm{magnetic}\:\mathrm{moment} \\ $$$$\mathrm{of}\:_{\mathrm{25}} \mathrm{Mn}^{\mathrm{x}+} \:\mathrm{is}\:\sqrt{\mathrm{15}}\mathrm{B}.\mathrm{M}.\:\mathrm{Then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{is}? \\ $$$$\mathrm{i}\:\mathrm{did}\:\mathrm{following} \\ $$$$\mathrm{1s}^{\mathrm{2}} \mathrm{2s}^{\mathrm{2}} \mathrm{2p}^{\mathrm{6}} \mathrm{3s}^{\mathrm{2}} \mathrm{3p}^{\mathrm{6}} \mathrm{4s}^{\mathrm{2}}…

Question-148276

Question Number 148276 by 0731619 last updated on 26/Jul/21 Answered by mathmax by abdo last updated on 26/Jul/21 $$\mathrm{y}\left(\mathrm{n}\right)=\mathrm{e}^{\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}} }\mathrm{log2}} \:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dn}}=\frac{\mathrm{d}}{\mathrm{dn}}\left(\mathrm{log2}×\frac{\mathrm{n}^{\mathrm{2}} \:+\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}}…