Question Number 160895 by HongKing last updated on 08/Dec/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}-\mathrm{1}} \:+\:…\:+\:\mathrm{1}}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:\left[\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{n}}\right)\:-\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right] \\ $$ Answered by Kamel last…
Question Number 160894 by HongKing last updated on 08/Dec/21 Answered by Kamel last updated on 09/Dec/21 $${a}_{{n}+\mathrm{1}} ={a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{3}} +{a}_{{n}} −{a}_{{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{\mathrm{2021}}…
Question Number 29820 by Tinkutara last updated on 12/Feb/18 Commented by ajfour last updated on 12/Feb/18 $$\left(\mathrm{1}\right)\:{in}\:{G}.{P}. \\ $$ Commented by math solver last updated…
Question Number 160883 by cortano last updated on 08/Dec/21 $$\:\sqrt{\mathrm{x}+\mathrm{1}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{2x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2x}+\mathrm{1}}\:−\mathrm{3}\:}\: \\ $$$$\:\mathrm{x}\:\in\mathbb{R}\: \\ $$ Answered by MJS_new last updated on 08/Dec/21 $$\mathrm{this}\:\mathrm{looks}\:\mathrm{suspicious}\:\mathrm{to}\:\mathrm{me}\:\mathrm{because} \\ $$$$\sqrt{\varphi+\mathrm{1}}=\left(\mathrm{2}\varphi+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}}…
Question Number 160879 by amin96 last updated on 08/Dec/21 Commented by cortano last updated on 08/Dec/21 $$\Leftrightarrow\:\mathrm{27}.\mathrm{x}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)} \:=\:\mathrm{x}^{\frac{\mathrm{10}}{\mathrm{3}}} \\ $$$$\Leftrightarrow\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{27}\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} =\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{log}\:_{\mathrm{3}}…
Question Number 29805 by mrW2 last updated on 12/Feb/18 $${f}\left({x}\right)=\left({x}+{a}_{\mathrm{1}} \right)\left({x}+{a}_{\mathrm{2}} \right)\left({x}+{a}_{\mathrm{3}} \right)…\left({x}+{a}_{{n}} \right) \\ $$$${find}\:{the}\:{coefficient}\:{of}\:{term}\:{x}^{{k}} \:\left(\mathrm{0}\leqslant{k}\leqslant{n}\right) \\ $$ Commented by mrW2 last updated on…
Question Number 160869 by HongKing last updated on 08/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\left(\mathrm{x}\:+\:\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:\mathrm{dx}\:=\:? \\ $$ Commented by smallEinstein last updated on 10/Dec/21 Commented…
Question Number 160871 by cortano last updated on 08/Dec/21 $$\:\:\sqrt[{\mathrm{6}}]{\left(\mathrm{2}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{13}}+\mathrm{5}}{\:\sqrt{\mathrm{5}}+\mathrm{2}}}\:+\mathrm{2}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{13}}−\mathrm{5}}{\:\sqrt{\mathrm{5}}−\mathrm{2}}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}=? \\ $$ Answered by MJS_new last updated on 08/Dec/21 $$\left(\pm\mathrm{5}+\mathrm{2}\sqrt{\mathrm{13}}\right)^{\mathrm{1}/\mathrm{3}} =\pm\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\left(\pm\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\pm\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}…
Question Number 160867 by cortano last updated on 08/Dec/21 $$ \\ $$A piece of metal in the form of an equilateral triangle that has been…
Question Number 95328 by mr W last updated on 24/May/20 Commented by mr W last updated on 24/May/20 $${reposted}\:{question} \\ $$ Commented by MJS last…