Question Number 212108 by Ismoiljon_008 last updated on 01/Oct/24 $$ \\ $$$$\:\:\:\int_{\:−\mathrm{1}} ^{\:\:\mathrm{1}} \mid\:{x}\:\mid\:\centerdot\:{ln}\left({x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)\:{dx}\:=\:? \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:{please} \\ $$$$ \\ $$ Answered by Frix last…
Question Number 212107 by MrGaster last updated on 01/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }{dx}. \\ $$$$ \\ $$ Answered by Frix last updated on…
Question Number 212114 by Ismoiljon_008 last updated on 01/Oct/24 $$ \\ $$$$\:\:\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$$\:\:\:\mathcal{F}{ind}\:{that}:\:\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:? \\ $$$$ \\ $$ Answered by Frix…
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Question Number 212097 by liuxinnan last updated on 30/Sep/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}^{\mathrm{2}} +{e}^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$ Answered by mehdee7396 last updated on 30/Sep/24 $${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{2}}…
Question Number 212098 by CrispyXYZ last updated on 30/Sep/24 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{ln}\:\frac{\sqrt{\mathrm{13}}−\mathrm{1}}{\mathrm{10}}\:+\:\sqrt{\mathrm{13}}\:−\:\mathrm{2}\:>\mathrm{0} \\ $$$$\mathrm{without}\:\mathrm{calculator}. \\ $$ Answered by MrGaster last updated on 03/Nov/24 $$\mathrm{ln}\left(\sqrt{\mathrm{13}}−\mathrm{1}\right)−\mathrm{ln}\:\mathrm{10}+\sqrt{\mathrm{13}}−\mathrm{2}>\mathrm{0} \\…
Question Number 212099 by vahid last updated on 30/Sep/24 Answered by mehdee7396 last updated on 30/Sep/24 $$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}\:\:\:\:\:;\:{let}\:\:\:{tan}\frac{{x}}{\mathrm{2}}={u} \\ $$$$=\int\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}={ln}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}+{c} \\ $$$$={ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\right)+{c}\:\:\checkmark \\…
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Question Number 212094 by behi834171 last updated on 29/Sep/24 $$\left[\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}+…….+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1000000}}}\right]=? \\ $$$$\boldsymbol{{note}}:\:\:\:\left[\mathrm{6}.\mathrm{25}\right]=\mathrm{6}\:\:\:,\left[\mathrm{0}.\mathrm{47}\right]=\mathrm{0} \\ $$ Answered by fabricio2008 last updated on 30/Sep/24 $$\underset{{x}=\mathrm{1}} {\overset{\mathrm{10}^{\mathrm{6}} } {\sum}}\left(\sqrt{{x}}\right)^{-\mathrm{1}}…
Question Number 212084 by Spillover last updated on 28/Sep/24 Terms of Service Privacy Policy Contact: info@tinkutara.com