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Category: Integration

calculate-dx-x-2-1-x-2-2-x-2-3-1-find-the-value-of-0-dx-x-2-1-x-2-2-x-2-3-

Question Number 39019 by maxmathsup by imad last updated on 01/Jul/18 $${calculate}\:\int\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)} \\ $$ Commented…

find-2x-3-x-2-x-3-8-dx-2-calculate-1-2x-3-x-2-x-3-8-dx-

Question Number 39017 by maxmathsup by imad last updated on 01/Jul/18 $${find}\:\:\:\int\:\:\frac{−\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left(\:{x}^{\mathrm{3}} \:+\mathrm{8}\right)}{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{−\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} \:+\mathrm{8}\right)}{dx} \\ $$ Commented by math…

find-dx-x-2x-1-3x-2-2-calculate-1-2-dx-x-2x-1-3x-2-

Question Number 39015 by maxmathsup by imad last updated on 01/Jul/18 $${find}\:\:\int\:\:\:\:\:\:\frac{{dx}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{3}{x}+\mathrm{2}\right)} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dx}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{3}{x}+\mathrm{2}\right)} \\ $$ Commented by math khazana by abdo last…

dx-Acos-x-B-

Question Number 104459 by bemath last updated on 21/Jul/20 $$\int\:\frac{{dx}}{\:\sqrt{{A}\mathrm{cos}\:{x}+{B}}} \\ $$ Commented by Dwaipayan Shikari last updated on 21/Jul/20 $$\int\frac{{dx}}{\:\sqrt{{Acosx}+{B}}}=\int\frac{{A}\left(−{sinx}\right)}{−{Asinx}}.\frac{\mathrm{1}}{\:\sqrt{{Acosx}+{B}}}{dx}\:\:\left\{{Acosx}+{B}={t}^{\mathrm{2}} \right. \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}\left(−{Asinx}\right)}=−\frac{\mathrm{2}}{{A}}\int\frac{\mathrm{1}}{{sinx}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{cosx}=\left({t}^{\mathrm{2}} −{B}\right).\frac{\mathrm{1}}{{A}}\right.…