Question Number 147066 by tabata last updated on 17/Jul/21 $${find}\:{laurant}\:{series}\:{lf} \\ $$$$ \\ $$$$\:\left(\mathrm{1}\right){f}\left({z}\right)=\frac{\mathrm{1}}{{z}−\mathrm{1}}+\frac{\mathrm{1}}{{z}+\mathrm{2}}\:\:,\mid{z}\mid>\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{2}\right){f}\left({z}\right)=\frac{\mathrm{1}}{{z}−\mathrm{2}}−\frac{\mathrm{2}}{{z}−\mathrm{3}}\:\:,\mid{z}\mid<\mathrm{3} \\ $$ Answered by mathmax by abdo…
Question Number 15972 by icyfalcon999 last updated on 16/Jun/17 $$\int\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}\: \\ $$ Commented by tawa tawa last updated on 16/Jun/17 $$\int\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:\mathrm{secx}\:\mathrm{dx} \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{trigonometry}\:\mathrm{identity}:\:\mathrm{tan}^{\mathrm{2}}…
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Question Number 146994 by KONE last updated on 17/Jul/21 $${u}_{{n}} ={cos}\sqrt{{n}+\mathrm{1}}−{cos}\sqrt{{n}} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{u}_{{n}} =?? \\ $$ Answered by mathmax by abdo last updated on…
Question Number 146987 by KONE last updated on 16/Jul/21 Answered by Olaf_Thorendsen last updated on 17/Jul/21 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{Supposons}\:\mathrm{que}\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{on}\:\:\mathrm{a}\:\mathrm{donc}\:\mathrm{les}\:\mathrm{solutions}\:\mathrm{suivantes}\:: \\ $$$$\left(\mathrm{0},\mathrm{0},{n}\right) \\ $$$$\left(\mathrm{0},\mathrm{1},{n}−\mathrm{1}\right)…
Question Number 146977 by tabata last updated on 16/Jul/21 $${find} \\ $$$$\left(\mathrm{1}\right)\:\int_{{C}} \:\frac{{e}^{{z}^{\mathrm{2}} } }{{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{3}}{dz}\:\:,{C}:\mid{z}−\mathrm{2}\mid=\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int_{−\infty} ^{\:\infty} \frac{{cosx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$…
Question Number 146975 by tabata last updated on 16/Jul/21 $${find}\:{laurant}\:{series}\:{f}\left({z}\right)=\frac{{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{3}}{{z}−\mathrm{2}}\:,\mid{z}−\mathrm{1}\mid>\mathrm{1} \\ $$ Answered by mathmax by abdo last updated on 17/Jul/21 $$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{3}}{\mathrm{z}−\mathrm{2}}=\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{2z}}{\mathrm{z}−\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{z}−\mathrm{2}}=\mathrm{z}\:+\frac{\mathrm{3}}{\mathrm{z}−\mathrm{2}}…
Question Number 81435 by naka3546 last updated on 13/Feb/20 Commented by naka3546 last updated on 13/Feb/20 $$\mathrm{3},\:\mathrm{4},\:\mathrm{6}\:\:{are}\:\:{Area}\:\:{of}\:\:{three}\:\:{triangles}\:\:{respectively}\:. \\ $$$${Shaded}\:\:{area}\:\:{is}\:\:…\: \\ $$ Commented by john santu…
Question Number 146959 by Gbenga last updated on 16/Jul/21 $$\int{ln}\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\right){dx} \\ $$ Answered by mindispower last updated on 17/Jul/21 $$\Leftrightarrow\int{ln}\left(\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}}\right){dy} \\ $$$$\left.\Leftrightarrow{yln}\left(\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}}\right)\right)−\int\frac{\mathrm{2}{y}^{\mathrm{2}}…