Question Number 130594 by Lordose last updated on 27/Jan/21 $$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx} \\ $$ Answered by mindispower last updated on 27/Jan/21 $$\left.\:\:{existe}\:{if}\:{n}\in\right]−\mathrm{1},\mathrm{5}\left[=\int_{\mathrm{0}} ^{\infty}…
Question Number 65061 by mathmax by abdo last updated on 24/Jul/19 $${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({x}−{t}\:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{with}\:\:\:{x}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{also}\:\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({x}−{t}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} } \\…
Question Number 65059 by mathmax by abdo last updated on 24/Jul/19 $${calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$ Commented by ~ À ® @ 237…
Question Number 130560 by rs4089 last updated on 26/Jan/21 Answered by Dwaipayan Shikari last updated on 26/Jan/21 $$\frac{\pi}{\mathrm{2}\left({r}+{r}^{\mathrm{3}} +{r}^{\mathrm{6}} +{r}^{\mathrm{10}} +….\right)} \\ $$ Answered by…
Question Number 130556 by Ar Brandon last updated on 26/Jan/21 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{xcosx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 65015 by aliesam last updated on 24/Jul/19 $$\int\frac{\sqrt{{x}+\mathrm{1}}\:−\:\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:+\:\sqrt{{x}−\mathrm{1}}}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 24/Jul/19 $${let}\:{I}\:=\int\:\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}}{dx}\:\Rightarrow{I}\:=\int\:\frac{\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} }{{x}+\mathrm{1}−{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}^{\mathrm{2}}…
Question Number 130544 by EDWIN88 last updated on 26/Jan/21 $$\:\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:+\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\mathrm{tan}\:{x}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)}\:{dx}? \\ $$ Answered by mindispower last updated on 26/Jan/21…
Question Number 65003 by mathmax by abdo last updated on 24/Jul/19 $${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\Gamma\left({x}\right)}\:\:{with}\:\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:\:\left({x}>\mathrm{0}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 65004 by mathmax by abdo last updated on 24/Jul/19 $${let}\:{U}_{{n}} =\:\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{2}}{{n}}} \:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right){dx}\:\:\:\:{with}\:{n}\geqslant\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{and}\:{determine}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$ Commented by mathmax…
Question Number 130536 by mathmax by abdo last updated on 26/Jan/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\theta\:+\mathrm{1}\right)\mathrm{d}\theta \\ $$ Answered by Ar Brandon last updated on 26/Jan/21…