Question Number 212052 by Spillover last updated on 28/Sep/24 Answered by Ghisom last updated on 28/Sep/24 $$\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{M}\:\mathrm{ethod}\right] \\ $$$$=−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}= \\…
Question Number 212049 by Spillover last updated on 28/Sep/24 Answered by Ghisom last updated on 28/Sep/24 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:\sqrt{−\mathrm{ln}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{−\mathrm{ln}\:{x}}\right] \\ $$$$=−\mathrm{2}\underset{\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}}…
Question Number 212048 by Spillover last updated on 28/Sep/24 Answered by Ghisom last updated on 28/Sep/24 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right] \\…
Question Number 212085 by hardmath last updated on 28/Sep/24 $$\mathrm{a},\mathrm{b},\mathrm{c}\:\in\:\mathbb{N} \\ $$$$\mathrm{5a}\:+\:\mathrm{6b}\:+\:\mathrm{7c}\:=\:\mathrm{70} \\ $$$$\mathrm{find}:\:\:\mathrm{max}\left(\mathrm{a}\right)\:=\:? \\ $$ Commented by Frix last updated on 28/Sep/24 $$\mathrm{If}\:\mathrm{0}\in\mathbb{N}\:\Rightarrow\:\mathrm{max}\:{a}\:=\mathrm{14}\:\:\:\:\:\left({b}={c}=\mathrm{0}\right) \\…
Question Number 212051 by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Ghisom last updated on 28/Sep/24 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}}…
Question Number 212083 by Spillover last updated on 28/Sep/24 Answered by Frix last updated on 28/Sep/24 $${I}=\int\:\frac{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\:−\mathrm{1}}{\mathrm{2}}{dx}= \\ $$$$=−\frac{{x}+\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{\mathrm{2}}+{C} \\ $$ Commented by Spillover last…
Question Number 212047 by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Terms of…
Question Number 212046 by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by…
Question Number 212075 by Spillover last updated on 28/Sep/24 Answered by Ghisom last updated on 28/Sep/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}+\sqrt{{n}}}−\sqrt{{n}}\right)\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{t}}}−\mathrm{1}}{\:\sqrt{{t}}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\sqrt{{t}}}}\:=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 212066 by Spillover last updated on 28/Sep/24 Answered by Rasheed.Sindhi last updated on 29/Sep/24 $$\frac{\mathrm{2021}!+\mathrm{2020}!}{\mathrm{2021}!−\mathrm{2020}!}\:\centerdot\:\frac{\mathrm{2020}!+\mathrm{2019}!}{\mathrm{2020}!−\mathrm{2019}!}\:\centerdot…\frac{\mathrm{3}!+\mathrm{2}!}{\mathrm{3}!−\mathrm{2}!}\:\centerdot\:\frac{\mathrm{2}!+\mathrm{1}!}{\mathrm{2}!−\mathrm{1}!} \\ $$$$\frac{\left\{\left({x}+\mathrm{1}\right)!+{x}!\right\}}{\left\{\left({x}+\mathrm{1}\right)!−{x}!\right\}}=\frac{{x}!\left\{\left({x}+\mathrm{1}\right)+\mathrm{1}\right\}}{{x}!\left\{\left({x}+\mathrm{1}\right)−\mathrm{1}\right\}}=\frac{{x}+\mathrm{2}}{{x}} \\ $$$$\frac{\mathrm{2020}+\mathrm{2}}{\mathrm{2020}}\centerdot\frac{\mathrm{2019}+\mathrm{2}}{\mathrm{2019}}\centerdot…\frac{\mathrm{2}+\mathrm{2}}{\mathrm{2}}\centerdot\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}} \\ $$$$\frac{\left(\mathrm{2020}+\mathrm{2}\right)\left(\mathrm{2019}+\mathrm{2}\right)\left(\mathrm{2018}+\mathrm{2}\right)…\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}\right)}{\mathrm{2020}.\mathrm{2019}.\mathrm{2018}….\mathrm{2}.\mathrm{1}} \\ $$$$=\frac{\mathrm{2022}.\mathrm{2021}.\mathrm{2020}….\mathrm{4}.\mathrm{3}}{\mathrm{2020}!}…