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Category: Integration

advanced-calculus-k-1-k-k-n-1-a-n-b-a-b-adapted-from-brilliant-k-0-1-

Question Number 137177 by mnjuly1970 last updated on 30/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:\:\:\:….\:\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\Phi=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} } \\ $$$$\:\:\:\:\:\:{a}\:,\:{b}\:=??\:\left({adapted}\:{from}\:{brilliant}\right) \\ $$$$\:\:\:\:\:\:\:……………. \\ $$$$\:\:\:\:\:\:\:\psi\left({k}\right)\overset{??} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}}…

x-x-3-x-2-dx-

Question Number 6107 by gourav~ last updated on 14/Jun/16 $$\int\frac{{x}}{{x}^{\mathrm{3}} +{x}+\mathrm{2}}{dx}\:=? \\ $$ Commented by Yozzii last updated on 14/Jun/16 $${x}^{\mathrm{3}} +{x}+\mathrm{2}={x}^{\mathrm{3}} +\mathrm{1}+{x}+\mathrm{1} \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}}…

Question-137171

Question Number 137171 by mnjuly1970 last updated on 30/Mar/21 Answered by Dwaipayan Shikari last updated on 30/Mar/21 $${log}\left(\mathrm{2}\right)=\mathrm{1}+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)+… \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}+..}}}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{1}+…}}}}} \\ $$$$\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}=\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}}…