Question Number 88010 by Chi Mes Try last updated on 07/Apr/20 $$\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$ Commented by niroj last updated on 07/Apr/20 $$\:\:\:\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}}{\:\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 22475 by mondodotto@gmail.com last updated on 19/Oct/17 Commented by Rasheed.Sindhi last updated on 19/Oct/17 $$\left(\mathrm{b}\right)\mathrm{9}^{\mathrm{8x}^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{9}^{\mathrm{16x}^{\mathrm{2}} } }=\left(\mathrm{9}^{\mathrm{16x}^{\mathrm{2}} } \right)^{\mathrm{1}/\mathrm{2}} =\mathrm{9}^{\mathrm{16x}^{\mathrm{2}}…
Question Number 22474 by Tinkutara last updated on 19/Oct/17 $$\mathrm{Let}\:{R}\:=\:\left(\mathrm{5}\sqrt{\mathrm{5}}\:+\:\mathrm{11}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{and}\:{f}\:=\:{R}\:−\:\left[{R}\right], \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{Rf}\:=\:\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22473 by Sahib singh last updated on 19/Oct/17 Commented by ajfour last updated on 19/Oct/17 $$\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{3}\right){are}\:{correct}. \\ $$$${For}\:{solution}\:{see}\:{Q}.\mathrm{22487} \\ $$ Terms of Service…
Question Number 22472 by Tinkutara last updated on 18/Oct/17 $$\mathrm{If}\:{x}^{{x}} \centerdot{y}^{{y}} \centerdot{z}^{{z}} \:=\:{x}^{{y}} \centerdot{y}^{{z}} \centerdot{z}^{{x}} \:=\:{x}^{{z}} \centerdot{y}^{{x}} \centerdot{z}^{{y}} \:\mathrm{such} \\ $$$$\mathrm{that}\:{x},\:{y}\:\mathrm{and}\:{z}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{1},\:\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}…
Question Number 88007 by Mikael_786 last updated on 07/Apr/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}=? \\ $$ Answered by MJS last updated on 07/Apr/20 $${t}=\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} \sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dt} \\ $$$$\Rightarrow…
Question Number 88004 by M±th+et£s last updated on 07/Apr/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\mathrm{2}{n}+\mathrm{1}}\left(\pi^{\mathrm{2}} +\mathrm{2}{H}_{{n}} ^{\left(\mathrm{2}\right)} −\mathrm{8}{H}_{\mathrm{2}{n}} ^{\left(\mathrm{2}\right)} \right) \\ $$$$=\frac{\mathrm{83}}{\mathrm{4}}\zeta\left(\mathrm{4}\right)−\mathrm{7}{log}\left(\mathrm{2}\right)\zeta\left(\mathrm{3}\right)−\mathrm{8}{log}^{\mathrm{2}} \left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right)−\frac{\mathrm{2}}{\mathrm{3}}{log}^{\mathrm{4}} \left(\mathrm{2}\right)−\mathrm{16}\:{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${where}\:{H}_{{n}}…
Question Number 153542 by EDWIN88 last updated on 08/Sep/21 $$\:{y}'''+{y}'=\mathrm{sec}\:{x}\: \\ $$ Answered by puissant last updated on 08/Sep/21 $${y}'''+{y}'={secx} \\ $$$$\Rightarrow\:{y}''+{y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}} \\ $$$$\Rightarrow\:\left({D}^{\mathrm{2}} +{D}\right){y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}}…
Question Number 22468 by Tinkutara last updated on 18/Oct/17 $$\mathrm{If}\:{a}_{{r}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\left(\mathrm{1}\:+\:{x}\:+\:{x}^{\mathrm{2}} \right)^{{n}} ,\:\mathrm{then} \\ $$$${a}_{\mathrm{1}} \:−\:\mathrm{2}{a}_{\mathrm{2}} \:+\:\mathrm{3}{a}_{\mathrm{3}} \:−\:…….\:\mathrm{2}{na}_{\mathrm{2}{n}} \:= \\ $$ Terms…
Question Number 153537 by SANOGO last updated on 08/Sep/21 Answered by puissant last updated on 08/Sep/21 $$\left.\mathrm{18}\right) \\ $$$${on}\:{remarque}\:{d}'{abord}\:{que} \\ $$$${x}^{{x}} =\mathrm{1}+{xlnx}+….+\frac{\left({xlnx}\right)^{{n}} }{{n}!}\left(\mathrm{1}+\varepsilon\left({x}\right)\right) \\ $$$$\varepsilon\left({x}\right)\rightarrow\mathrm{0}\:{quand}\:{x}\rightarrow\mathrm{0}…