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Category: Integration

find-ln-x-x-2-1-x-2-1-dx-2-calculate-2-5-ln-x-x-2-1-x-2-1-dx-

Question Number 39833 by math khazana by abdo last updated on 12/Jul/18 $${find}\:\:\int\:\:\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\frac{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right.}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$ Commented by…

calculste-I-cos-x-n-1-x-2-dx-with-from-R-and-n-integr-natural-2-find-the-vslue-of-cos-3-x-9-1-x-2-dx-

Question Number 39787 by abdo mathsup 649 cc last updated on 10/Jul/18 $${calculste}\:\:{I}_{\lambda} \:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\lambda{x}^{{n}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{with} \\ $$$$\lambda\:{from}\:{R}\:{and}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{vslue}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{3}\:{x}^{\mathrm{9}} \right)}{\mathrm{1}+{x}^{\mathrm{2}}…

Question-170831

Question Number 170831 by venom1 last updated on 01/Jun/22 Answered by LEKOUMA last updated on 01/Jun/22 $$\left.\mathrm{1}\right)\:\int\mathrm{4sin}\:\mathrm{8}{xdx}=\mathrm{4}\int\mathrm{sin}\:\mathrm{8}{xdx}=\mathrm{4}×−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{8}{x}+{c}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{8}{x}+{c} \\ $$$$\left.\mathrm{2}\right)\:\int{x}\mathrm{cos}\:\left(\mathrm{12}{x}^{\mathrm{2}} \right){dx} \\ $$$${let}\:{u}=\mathrm{12}{x}^{\mathrm{2}} \:\Rightarrow\:{du}=\mathrm{24}{xdx}\:\Rightarrow\:{dx}=\frac{{du}}{\mathrm{24}{x}} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{24}}\mathrm{cos}\left(\:{u}\right){du}=\frac{\mathrm{1}}{\mathrm{24}}\int\mathrm{cos}\:\left({u}\right){du}=\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}\left(\:{u}\right)+{c}…

Question-170802

Question Number 170802 by Sotoberry last updated on 31/May/22 Answered by thfchristopher last updated on 31/May/22 $$\int\mathrm{ln}\:\left({x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$=\int\mathrm{ln}\:{x}\left({x}+\mathrm{1}\right){dx} \\ $$$$=\int\mathrm{ln}\:{xdx}+\int\mathrm{ln}\:\left({x}+\mathrm{1}\right){dx} \\ $$$$={x}\mathrm{ln}\:{x}−\int{xd}\left(\mathrm{ln}\:{x}\right)+{x}\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\int{xd}\left[\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right] \\…