Question Number 41627 by alex041103 last updated on 10/Aug/18 Commented by alex041103 last updated on 10/Aug/18
Question Number 107162 by bobhans last updated on 09/Aug/20 Commented by Dwaipayan Shikari last updated on 09/Aug/20 $$\int_{−\mathrm{5}} ^{\mathrm{5}} \sqrt[{\mathrm{6}}]{\mathrm{5x}^{\mathrm{2}} +\mathrm{6}}\:\left(\mathrm{5x}−\mathrm{2}\right)\mathrm{dx}=\int_{−\mathrm{5}} ^{\mathrm{5}} \sqrt[{\mathrm{6}}]{\mathrm{5x}^{\mathrm{2}} +\mathrm{6}}\:\left(−\mathrm{5x}−\mathrm{2}\right)=\mathrm{I} \…
Question Number 107141 by cesarL last updated on 09/Aug/20
Question Number 41586 by MJS last updated on 09/Aug/18
Question Number 172653 by Mikenice last updated on 29/Jun/22 Answered by MJS_new last updated on 30/Jun/22
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Question Number 107107 by mohammad17 last updated on 08/Aug/20 Commented by Dwaipayan Shikari last updated on 08/Aug/20 $$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{y}}\:\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}^{\mathrm{3}} =\mathrm{t}^{\mathrm{2}\:\:} \:,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{t}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{y}}\:\:=\frac{\mathrm{dt}}{\mathrm{dy}}…
Question Number 172636 by Mikenice last updated on 29/Jun/22 Answered by floor(10²Eta[1]) last updated on 29/Jun/22 $$\mathrm{I}=\int\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}−\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \…
Question Number 41561 by Tawa1 last updated on 09/Aug/18