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Category: Integration

dx-x-2-3-1-x-2-3-

Question Number 36597 by rahul 19 last updated on 03/Jun/18 $$\int\frac{\mathrm{d}{x}}{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}+{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}\:=\:? \\ $$ Commented by rahul 19 last updated on 03/Jun/18 $$\mathrm{I}\:\mathrm{tried}\:\mathrm{by}\:\mathrm{taking}\:{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{common}\:{from}…

x-1-x-1-dx-

Question Number 102128 by Study last updated on 06/Jul/20 $$\:\int\frac{{x}+\mathrm{1}}{\:\sqrt{{x}}\:+\mathrm{1}}{dx}=? \\ $$ Answered by Dwaipayan Shikari last updated on 06/Jul/20 $$\int\frac{{x}}{\:\sqrt{{x}}+\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}}=\int\frac{\mathrm{2}{u}^{\mathrm{3}} }{{u}+\mathrm{1}}{du}+\int\frac{\mathrm{2}{u}}{{u}+\mathrm{1}}{du}\:\:\:\:\left\{\:\:\:{take}\:{x}\:\:{as}\:{u}^{\mathrm{2}} \right. \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{3}}…

s-s-0-x-s-1-e-x-1-dx-Prove-that-And-prove-1-2-3-4-5-6-7-1-12-

Question Number 102115 by Dwaipayan Shikari last updated on 06/Jul/20 $$\Gamma\left({s}\right)\zeta\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} +\mathrm{1}}{dx}\:\:\left({Prove}\:{that}\right) \\ $$$${And}\:{prove}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+….\infty=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$ Commented by mr W…

Question-167647

Question Number 167647 by mathlove last updated on 22/Mar/22 Answered by RoswelCod2003 last updated on 22/Mar/22 $$\mathrm{Using}\:“\mathrm{Feynman}'\mathrm{s}\:\mathrm{Rule}\:\mathrm{of}\:\mathrm{Integration}'' \\ $$$$\: \\ $$$${I}\left({k}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx}\:\therefore\:\mathrm{where}\:{k}\:=\:\mathrm{2}…

sinx-cosx-cos2x-cos4x-dx-

Question Number 102099 by Study last updated on 06/Jul/20 $$\int{sinx}\:\centerdot\:{cosx}\:\centerdot{cos}\mathrm{2}{x}\:\centerdot\:{cos}\mathrm{4}{x}\:{dx}=? \\ $$ Answered by PRITHWISH SEN 2 last updated on 06/Jul/20 $$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{2xcos}\:\mathrm{2xcos}\:\mathrm{4x}\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sin}\:\mathrm{4xcos}\:\mathrm{4xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{sin}\:\:\mathrm{8xdx}=−\frac{\mathrm{cos}\:\:\mathrm{8x}}{\mathrm{64}}\:+\mathrm{C} \\…