Question Number 223078 by Nicholas666 last updated on 14/Jul/25 $$ \\ $$$$\:\:\:\:\:\:\mathrm{Evaluate}\::\:\int\:\frac{\mathrm{ln}\:{x}\:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}\:\mathrm{d}{x}\: \\ $$$$ \\ $$ Answered by MrGaster last updated on 14/Jul/25 Terms…
Question Number 223079 by Nicholas666 last updated on 14/Jul/25 $$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{Evaluate}}\::\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left({x}\right)}{{x}}\:\mathrm{Li}_{\mathrm{2}} \left({x}\right)\:\mathrm{d}{x} \\ $$$$ \\ $$ Answered by MrGaster last updated on…
Question Number 223032 by MrGaster last updated on 13/Jul/25 $$ \\ $$$$\:\:\:\:\underset{{t}=\mathrm{0}} {\overset{\mathrm{2}^{\mathrm{2024}} } {\prod}}\left(\mathrm{4sin}^{\mathrm{2}} \frac{{t}\pi}{\mathrm{2}^{\mathrm{2025}} }−\mathrm{3}\right) \\ $$ Answered by MrGaster last updated on…
Question Number 223066 by fantastic last updated on 13/Jul/25 Commented by fantastic last updated on 13/Jul/25 $${Area}\:{of}\:{semicircle}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b} \\ $$$$ \\ $$ Answered by mr W…
Question Number 223034 by MrGaster last updated on 13/Jul/25 $$ \\ $$$$\int\frac{{x}\:\mathrm{arctan}{xdx}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{k}'^{\mathrm{2}} {x}^{\mathrm{2}} \right)}}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left[{F}\left(\frac{\pi}{\mathrm{4}},{k}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+{k}'\mathrm{2}\right)}}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223054 by Silver last updated on 13/Jul/25 $$\mathrm{find}\:{y} \\ $$$${y}^{{dy}} =\:{x}^{{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223055 by alcohol last updated on 13/Jul/25 Commented by alcohol last updated on 13/Jul/25 $${please}\:{i}\:{really}\:{really}\:{need}\:{a}\:{detailed} \\ $$$${explanation} \\ $$ Commented by mr W…
Question Number 223049 by wewji12 last updated on 13/Jul/25 $$\mathrm{we}\:\mathrm{already}\:\mathrm{know}\:\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\infty\:\mathrm{and}\:\underset{{p}\in\mathbb{P}} {\sum}\:\frac{\mathrm{1}}{{p}}=\infty \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{Harmonic}\:\mathrm{Series} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\underset{{p}\in\mathbb{P}} {\prod}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}+\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{3}} +…\right)=\:\underset{{p}\in\mathbb{P}} {\prod}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−\mathrm{1}} } \\…
Question Number 223044 by hardmath last updated on 13/Jul/25 Commented by hardmath last updated on 13/Jul/25 $$ \\ $$Find the voltmeter reading in the electrical…
Question Number 223040 by MrGaster last updated on 13/Jul/25 Commented by Nicholas666 last updated on 14/Jul/25 $$\:\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir},\: \\ $$$$\:\mathrm{very}\:\mathrm{nice}\:\mathrm{you}'\mathrm{re}\:\mathrm{approach}\:\:\:\mathrm{solution}\:\mathrm{to}\:\mathrm{this}\:\mathrm{problem}\: \\ $$$$ \\ $$ Terms of…