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Category: Integration

Question-219549

Question Number 219549 by Spillover last updated on 28/Apr/25 Answered by Nicholas666 last updated on 29/Apr/25 1+sinx1sinxdx=2tanyx+c(1+sinx)21sin2xdx=2tanyx+c$$\int\frac{\mathrm{1}+\mathrm{2}\:{sin}\:{x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}+{c}…

I-n-0-1-0-1-0-1-ln-1-x-1-x-2-x-n-1-x-1-1-x-2-1-x-n-dx-1-dx-2-dx-n-

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}}…