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x-3-1-x-2-dx-




Question Number 140676 by ZiYangLee last updated on 11/May/21
∫ (x^3 /(1+x^2 )) dx=?
$$\int\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}=? \\ $$
Answered by Dwaipayan Shikari last updated on 11/May/21
∫(x^3 /(1+x^2 ))dx  =∫x−(x/(1+x^2 ))dx=(x^2 /2)−(1/2)log(1+x^2 )+C
$$\int\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int{x}−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{C} \\ $$
Answered by mathmax by abdo last updated on 11/May/21
∫  (x^3 /(x^2  +1))dx =∫  ((x(x^2 +1)−x)/(x^2  +1))dx  =∫ xdx−∫  (x/(x^2  +1))dx =(x^2 /2)−(1/2)log(x^2  +1)+C
$$\int\:\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\int\:\mathrm{xdx}−\int\:\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{C} \\ $$

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