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Question Number 227168 by mr W last updated on 04/Jan/26 $${tinku}\:{tarra}\:{sir}: \\ $$$${can}\:{you}\:{please}\:{check}\:{following}\:{issue}: \\ $$$${this}\:{is}\:{a}\:{table}\:{with}\:{frame}: \\ $$$$\begin{array}{|c|c|}{\mathrm{1}}&\hline{\mathrm{10}}\\{\mathrm{2}}&\hline{\mathrm{20}}\\\hline\end{array} \\ $$$${this}\:{is}\:{a}\:{table}\:{without}\:{frame}: \\ $$$$\begin{matrix}{\mathrm{1}}&{\mathrm{10}}\\{\mathrm{2}}&{\mathrm{20}}\end{matrix} \\ $$$${but}\:{when}\:{this}\:{post}\:{is}\:{uploaded},\:{the} \\ $$$${second}\:{table}\:{and}\:{all}\:{things}\:…

Question-226936

Question Number 226936 by Estevao last updated on 19/Dec/25 Answered by Ghisom_ last updated on 19/Dec/25 $$\int\mathrm{2}^{\mathrm{ln}\:{x}} {dx}=\int{x}^{\mathrm{ln}\:\mathrm{2}} {dx}=\frac{{x}^{\mathrm{1}+\mathrm{ln}\:\mathrm{2}} }{\mathrm{1}+\mathrm{ln}\:\mathrm{2}}+{C} \\ $$ Commented by Estevao…

Question-226914

Question Number 226914 by Estevao last updated on 18/Dec/25 Answered by breniam last updated on 19/Dec/25 $$\mathrm{We}\:\mathrm{can}\:\mathrm{concider}\:\mathrm{it}\:\mathrm{only}\:\mathrm{for}\:\mathrm{odd}\:{n},\:\mathrm{because} \\ $$$$\mathrm{real}\:\mathrm{even}\:\mathrm{root}\:\mathrm{of}\:\mathrm{negative}\:\mathrm{number}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}. \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{2}\right)!\left(\mathrm{2}−\mathrm{2}{n}\right)}= \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{1}\right)!}×\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\frac{\mathrm{2}−\mathrm{2}{n}}{\mathrm{2}{n}−\mathrm{1}}}…

Question-226866

Question Number 226866 by Estevao last updated on 17/Dec/25 Commented by Jyrgen last updated on 17/Dec/25 $$\int\sqrt{\mathrm{1}+{x}^{{n}} }{dx}={x}\:_{\mathrm{1}} {F}_{\mathrm{2}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{{n}};\:\mathrm{1}+\frac{\mathrm{1}}{{n}};\:−{x}^{{n}} \right) \\ $$ Commented by…