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Category: Integration

Question-168368

Question Number 168368 by SUPERMATH last updated on 09/Apr/22 Answered by qaz last updated on 09/Apr/22 $$\int\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\int\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}+\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\int\frac{\mathrm{x}+\mathrm{2x}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}}…

calculate-C-9-z-2-2-z-z-1-3-z-2-dz-with-C-is-the-circle-C-z-C-z-3-

Question Number 37299 by math khazana by abdo last updated on 11/Jun/18 $${calculate}\:\:\int_{{C}} \:\:\:\frac{\mathrm{9}\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)}{{z}\left({z}+\mathrm{1}\right)^{\mathrm{3}} \left({z}−\mathrm{2}\right)}{dz}\:\:{with}\:\:{C}\:{is}\:{the} \\ $$$${circle}\:{C}\:=\left\{{z}\in{C}/\:\mid{z}\mid\:=\mathrm{3}\right\}\: \\ $$ Commented by math khazana by…

Question-168366

Question Number 168366 by cortano1 last updated on 09/Apr/22 Answered by qaz last updated on 09/Apr/22 $$\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}=\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}}…