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Question-206522

Question Number 206522 by MrGHK last updated on 17/Apr/24 Answered by Berbere last updated on 17/Apr/24 $${u}'={xln}^{\mathrm{2}} \left({x}\right)\Rightarrow{u}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)−{x}^{\mathrm{2}} {ln}\left({x}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${v}={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{x}}{{x}}\right);{v}'=−\frac{\mathrm{1}}{{x}^{\mathrm{2}}…

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Question Number 206393 by MaruMaru last updated on 13/Apr/24 $$\mathrm{find}\:\mathrm{S}=\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right)\:,\:\ell\in\left[\mathrm{1},\infty\right) \\ $$$$\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right) \\ $$$$\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}\right)−…… \\ $$ Answered by MaruMaru last updated…

Question-206364

Question Number 206364 by Skabetix last updated on 12/Apr/24 Answered by TonyCWX08 last updated on 13/Apr/24 $${I}\:{only}\:{know} \\ $$$${e}^{\pi{i}} =−\mathrm{1} \\ $$$$\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:\left({x}\right)}{{x}}\:{dx}\:=\:\pi \\…

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Question Number 206351 by MetaLahor1999 last updated on 12/Apr/24 $${expression}\:{of}\:{the}\:{sequence}\:\left({a}_{{n}} \right)\:{defined} \\ $$$${by}\: \\ $$$$\begin{cases}{{a}_{\mathrm{0}} >\mathrm{0}\:,\:{a}_{\mathrm{1}} >\mathrm{0}}\\{{a}_{{n}+\mathrm{2}} =\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}−\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{3}\right)}{{n}+\mathrm{2}}{a}_{{n}+\mathrm{1}} +\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}{a}_{{n}} }\end{cases} \\ $$ Commented…