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Author: Tinku Tara

0-1-x-1-x-8-dx-

Question Number 137123 by bobhans last updated on 30/Mar/21 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\:\mathrm{dx}\:=? \\ $$ Commented by Ar Brandon last updated on 30/Mar/21 You're right, Sir. Greetings to you ! It's been quite a longtime since we last interracted. Haha ! Commented…

Question-137117

Question Number 137117 by MathZa last updated on 29/Mar/21 Answered by mathmax by abdo last updated on 30/Mar/21 $$\mathrm{is}\:\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\mathrm{0}? \\ $$$$\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{i}}\:+\sqrt{\mathrm{2}}\mathrm{i}\:=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}\:−\mathrm{i}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}}\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}\left(\mathrm{1}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}\right.}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}+\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }…

Question-137112

Question Number 137112 by Algoritm last updated on 29/Mar/21 Answered by Ñï= last updated on 03/Apr/21 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{9}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{23}}\right)+… \\ $$$$={arg}\left(\mathrm{3}+{i}\right)+{arg}\left(\mathrm{9}+\mathrm{2}{i}\right)+{arg}\left(\mathrm{23}+\mathrm{4}{i}\right)+… \\ $$$$={arg}\left[\left(\mathrm{3}+{i}\right)\left(\mathrm{9}+\mathrm{2}{i}\right)\left(\mathrm{23}+\mathrm{4}{i}\right)\centerdot…\right] \\…