Question Number 197784 by cortano12 last updated on 28/Sep/23 $$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$ Answered by Frix last updated on 28/Sep/23 $$\mathrm{Let}\:{x}={t}^{\mathrm{3}} +\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} +\mathrm{12}{t}^{\mathrm{2}}…
Question Number 197783 by pticantor last updated on 28/Sep/23 $$\int\frac{{x}.\boldsymbol{{arctg}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dx}}=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Answered by EmGent last…
Question Number 197776 by universe last updated on 28/Sep/23 Commented by witcher3 last updated on 28/Sep/23 $$\mid\underset{−−} {\mathrm{n}}\:\:\:\mathrm{what}\:\mathrm{is}\:? \\ $$ Commented by universe last updated…
Question Number 197772 by Erico last updated on 28/Sep/23 $$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{do}\:\mathrm{this}? \\ $$$$\underset{\:\mathrm{1}} {\int}^{\:+\infty} \frac{\mathrm{t}−\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{3}} \:\mathrm{lnt}}\mathrm{dt} \\ $$ Commented by witcher3 last updated on 28/Sep/23 $$\mathrm{we}\:\mathrm{can}\:\mathrm{Give}\:…
Question Number 197771 by sonukgindia last updated on 28/Sep/23 Answered by som(math1967) last updated on 28/Sep/23 $${B}=\frac{\mathrm{1}}{\mathrm{2}}×{r}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{45}}{\mathrm{360}}×\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right){squnit} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}}…
Question Number 197766 by universe last updated on 28/Sep/23 $$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\mathrm{sin}^{\mathrm{2}{n}+\mathrm{1}} {x}\:\mathrm{cos}\:{x}\right){dx}\:\:=\:? \\ $$ Commented by witcher3 last updated on 28/Sep/23 $$\mathrm{dominate}\:\mathrm{cv}\:\mathrm{Theorem} \\…
Question Number 197767 by AR19 last updated on 28/Sep/23 Commented by Frix last updated on 28/Sep/23 $$\int\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}{dx}={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:−{x}^{\mathrm{3}} \right)\:+{C} \\ $$ Terms of…
Question Number 197792 by Mastermind last updated on 28/Sep/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{3}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}''\:+\:\mathrm{y}^{'} \:−\:\mathrm{2y}\:=\:\mathrm{x}\:+\:\mathrm{sin2x},\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{y}''\:−\:\mathrm{y}'\:=\:\mathrm{xe}^{\mathrm{x}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$ \\…
Question Number 197794 by universe last updated on 28/Sep/23 $$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$ Answered by Frix last updated on 29/Sep/23 $${x}=\mathrm{ln}\:\mathrm{tan}\:\frac{\mathrm{2}{y}+\pi}{\mathrm{4}}\:\Leftrightarrow\:{y}=−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \\ $$$$−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}}…
Question Number 197763 by luciferit last updated on 28/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com