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Category: Integration

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}}…

Question-137139

Question Number 137139 by peter frank last updated on 30/Mar/21 Answered by Dwaipayan Shikari last updated on 30/Mar/21 $$\int_{\mathrm{1}} ^{{e}} \frac{\mathrm{1}}{\left(\mathrm{1}+{log}\left({x}\right)\right)}−\frac{\mathrm{1}}{\left(\mathrm{1}+{log}\left({x}\right)\right)^{\mathrm{2}} }{dx}\:\:\:\:\:{log}\left({x}\right)={t} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}}…

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-137111

Question Number 137111 by mnjuly1970 last updated on 29/Mar/21 Answered by Dwaipayan Shikari last updated on 29/Mar/21 $$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\left(\pi{coth}\left(\pi\right)−\mathrm{1}\right)=\pi\frac{{e}^{\mathrm{2}\pi} +\mathrm{1}}{{e}^{\mathrm{2}\pi}…