Category: Integration

Question Number 219549 by Spillover last updated on 28/Apr/25 Answered by Nicholas666 last updated on 29/Apr/25 $$\int\frac{\mathrm{1}+{sin}\:{x}}{\mathrm{1}−{sinx}}{dx}=\mathrm{2}{tany}−{x}+{c} \\ $$$$\int\frac{\left(\mathrm{1}+{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{1}−{sin}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}\:+{c} \\ $$$$\int\frac{\mathrm{1}+\mathrm{2}\:{sin}\:{x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}+{c}…

Question Number 219553 by Nicholas666 last updated on 28/Apr/25 $$ \\ $$$$\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} ….\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:….{x}_{{n}} \right)}{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)….\left(\mathrm{1}−{x}_{{n}\:} \right)}\:{dx}_{\mathrm{1}}…

Question Number 219509 by Nicholas666 last updated on 27/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{I}\:=\:\underset{\:\mathrm{1}} {\int}^{\:\mathrm{16}} \:\frac{\left({x}\:+\sqrt{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:{dx} \\ $$$$ \\ $$ Answered by SdC355 last updated…

Question Number 219529 by Nicholas666 last updated on 27/Apr/25 $$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\:\:\:\frac{\mathrm{1}+{x}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{x}^{\mathrm{4}} −{x}^{\mathrm{5}} }{\mathrm{1}−{x}^{\mathrm{7}} }\:\:\:{dx} \\ $$$$ \\ $$ Commented by…

Question Number 219482 by Nicholas666 last updated on 26/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{{e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$ \\ $$ Commented by Nicholas666 last updated on…