Question Number 125593 by shaker last updated on 12/Dec/20 Answered by MJS_new last updated on 12/Dec/20 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\left({search}\:{the}\:{www}\:{for}\:{it}\right)\right] \\ $$$$=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}=…
Question Number 60056 by bhanukumarb2@gmail.com last updated on 17/May/19 Commented by maxmathsup by imad last updated on 17/May/19 $${we}\:{see}\:{that}\:\:{S}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{sin}\left({k}\theta\right){cos}^{{k}} \theta}{{k}!}\:\Rightarrow{S}\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{ik}\theta} \:{cos}^{{k}}…
Question Number 60054 by bhanukumarb2@gmail.com last updated on 17/May/19 Commented by bhanukumarb2@gmail.com last updated on 17/May/19 $${is}\:{titu}\:{lemma}\:{applicable}\:{here}??{plz}\:{see} \\ $$$${previous}\:{doubts} \\ $$ Answered by tanmay last…
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Question Number 191126 by Mingma last updated on 18/Apr/23 Commented by mahdipoor last updated on 18/Apr/23 $$\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\frac{\mathrm{4}\left(\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ Terms of…
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Question Number 60050 by maxmathsup by imad last updated on 17/May/19 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$ Commented by maxmathsup by imad last updated…
Question Number 125585 by Dwaipayan Shikari last updated on 12/Dec/20 $$\frac{\left(\frac{\mathrm{1}}{\mathrm{1}!}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}!}\right)^{\mathrm{2}} +…\:}{\left(\frac{\mathrm{1}}{\mathrm{1}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} +……} \\ $$ Commented by Dwaipayan Shikari last…