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Author: Tinku Tara

prove-0-ln-x-1-x-4-ln-x-3-1-x-17-1-x-dx-240-3-2-

Question Number 215081 by MrGaster last updated on 28/Dec/24 $$ \\ $$$$\mathrm{prove}: \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{ln}\left({x}/\left(\mathrm{1}+{x}\right)\right)^{\mathrm{4}} \mathrm{ln}\left({x}^{\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{17}} \right.\right.}{\mathrm{1}+{x}}{dx}=−\mathrm{240}\zeta\left(\mathrm{3}\right)^{\mathrm{2}} \\ $$$$ \\ $$…

log-2-x-log-3-x-1-5-x-

Question Number 215089 by golsendro last updated on 28/Dec/24 $$\:\:\mathrm{log}\:_{\mathrm{2}} \:\mathrm{x}\:+\:\mathrm{log}\:_{\mathrm{3}} \:\left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{5}\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$ Answered by efronzo1 last updated on 28/Dec/24 $$\:\:\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{2}\:=\:\mathrm{3}−\mathrm{log}\:_{\mathrm{2}}…

n-0-1-n-2-n-1-2n-1-p-q-p-q-p-q-

Question Number 215091 by mnjuly1970 last updated on 28/Dec/24 $$ \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{\:\Gamma^{\:\mathrm{2}} \left({n}+\mathrm{1}\right)}{\Gamma\:\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\:? \\ $$$$\:\:\:\:\:\:−−− \\ $$$$\:\:\:\:\beta\:\left({p}\:,{q}\:\right)\:=\:\frac{\:\Gamma\:\left({p}\right)\Gamma\left({q}\right)}{\Gamma\left({p}+{q}\:\right)} \\ $$ Answered by MrGaster…

Question-215052

Question Number 215052 by Abdullahrussell last updated on 27/Dec/24 Answered by mr W last updated on 27/Dec/24 $${f}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right){g}\left({x}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right){g}\left(\mathrm{2}\right)+\mathrm{1}=\mathrm{4}\:\Rightarrow{g}\left(\mathrm{2}\right)=\mathrm{3}\:\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left[\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3}\right]+\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}−\mathrm{1}\right)\left[\left(\mathrm{3}−\mathrm{2}\right){h}\left(\mathrm{3}\right)+\mathrm{3}\right]+\mathrm{1}=\mathrm{3}\:\Rightarrow{h}\left(\mathrm{3}\right)=−\mathrm{2}\:\Rightarrow{h}\left({x}\right)=\left({x}−\mathrm{3}\right){k}\left({x}\right)−\mathrm{2} \\…

Solve-for-a-b-c-R-1-a-1-b-c-1-2-1-b-1-c-a-1-3-1-c-1-a-b-1-4-

Question Number 215048 by Hanuda354 last updated on 27/Dec/24 $$\mathrm{Solve}\:\:\mathrm{for}\:\:{a},\:{b},\:{c}\:\in\:\mathbb{R} \\ $$$$\:\:\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}+{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}+{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}+{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by Abdullahrussell last updated on 27/Dec/24…