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Author: Tinku Tara

V-ln-x-1-x-2-dx-

Question Number 133951 by liberty last updated on 25/Feb/21 $$\:\mathcal{V}\:=\:\int\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right)\:\mathrm{dx}\: \\ $$ Answered by mathmax by abdo last updated on 25/Feb/21 $$\Phi\:=\int\:\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{sht}\:\Rightarrow \\…

advanced-calculus-prove-that-0-cos-x-2-cos-x-x-dx-2-euler-mascheroni-constant-

Question Number 133943 by mnjuly1970 last updated on 25/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:\:\:\:\:{calculus}…… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({x}^{\mathrm{2}} \right)−{cos}\left({x}\right)}{{x}}{dx}=\frac{\gamma}{\mathrm{2}} \\ $$$$\:\:\:\gamma:\:{euler}−{mascheroni}\:{constant}… \\ $$ Terms of Service Privacy…

There-are-a-few-problem-reported-Notification-google-discontinued-Google-Cloud-Messaging-so-notifications-are-not-working-Other-Problems-There-has-been-several-new-phone-models-and-android-versio

Question Number 68405 by Tinku Tara last updated on 10/Sep/19 $$\mathrm{There}\:\mathrm{are}\:\mathrm{a}\:\mathrm{few}\:\mathrm{problem}\:\mathrm{reported}. \\ $$$$\mathrm{Notification}:\:\mathrm{google}\:\mathrm{discontinued} \\ $$$$\mathrm{Google}\:\mathrm{Cloud}\:\mathrm{Messaging}\:\mathrm{so}\:\mathrm{notifications} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{working}.\: \\ $$$$\mathrm{Other}\:\mathrm{Problems}: \\ $$$$\mathrm{There}\:\mathrm{has}\:\mathrm{been}\:\mathrm{several}\:\mathrm{new}\:\mathrm{phone} \\ $$$$\mathrm{models}\:\mathrm{and}\:\mathrm{android}\:\mathrm{version}\:\mathrm{updates}. \\ $$$$\mathrm{And}\:\mathrm{newer}\:\mathrm{android}\:\mathrm{version}\:\mathrm{or}\:\mathrm{phone}…

Question-133937

Question Number 133937 by rexford last updated on 25/Feb/21 Answered by EDWIN88 last updated on 25/Feb/21 $$\boldsymbol{{AB}}\:=\:\hat {\boldsymbol{\mathrm{i}}}+\mathrm{6}\hat {\boldsymbol{\mathrm{j}}}\:,\:\mathrm{let}\:\mathrm{vector}\:\boldsymbol{\mathrm{u}}\:=\:\boldsymbol{\mathrm{AC}}\:\mathrm{where}\:\mathrm{C}\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\boldsymbol{\mathrm{u}}=\left(\mathrm{x}−\mathrm{1},\mathrm{y}+\mathrm{1}\right)\:=\left(\mathrm{x}−\mathrm{1}\right)\hat {\boldsymbol{\mathrm{i}}}+\left(\mathrm{y}+\mathrm{1}\right)\hat {\boldsymbol{\mathrm{j}}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{u}}.\boldsymbol{\mathrm{AB}}\:=\mathrm{0}\:\Rightarrow\mathrm{x}−\mathrm{1}+\mathrm{6y}+\mathrm{6}\:=\:\mathrm{0}…

log-x-8-x-2-3x-4-lt-2-log-4-x-2-x-4-

Question Number 133936 by liberty last updated on 25/Feb/21 $$\mathrm{log}\:_{\mathrm{x}+\mathrm{8}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{4}\right)\:<\:\mathrm{2}.\mathrm{log}\:_{\left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{2}} } \left(\mid\mathrm{x}−\mathrm{4}\mid\right)\: \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{log}\:_{\mathrm{x}+\mathrm{8}} \left(\mathrm{x}^{\mathrm{2}}…