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Question Number 29574 by yesasitya22@gmail.com last updated on 10/Feb/18

$$\int x^6-1/x^2-1dx$$

Commented by tawa tawa last updated on 10/Feb/18

$$\int \frac{{\mathrm{x}}^6-1}{{\mathrm{x}}^2-1}\mathrm{dx}$$$$=\int \frac{{\left({\mathrm{x}}^2\right)}^3-1^3}{{\mathrm{x}}^2-1}\mathrm{dx}$$$$\mathrm{Simplify}\mathrm{the}\mathrm{numerator}\mathrm{using}\mathrm{the}\mathrm{identity}:{\mathrm{x}}^3-{\mathrm{y}}^3=(\mathrm{x}-\mathrm{y})({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2)$$$$=\int \frac{({\mathrm{x}}^2-1)[{\left({\mathrm{x}}^2\right)}^2+{\mathrm{x}}^2+1^2]}{{\mathrm{x}}^2-1}\mathrm{dx}$$$$=\int \frac{({\mathrm{x}}^2-1)({\mathrm{x}}^4+{\mathrm{x}}^2+1)}{{\mathrm{x}}^2-1}\mathrm{dx}$$$$=\int ({\mathrm{x}}^4+{\mathrm{x}}^2+1)\mathrm{dx}$$$$=\frac{{\mathrm{x}}^5}{5}+\frac{{\mathrm{x}}^3}{3}+\mathrm{x}+\mathrm{C}$$$$=\frac{1}{5}{\mathrm{x}}^5+\frac{1}{3}{\mathrm{x}}^3+\mathrm{x}+\mathrm{C}$$

Commented by ajfour last updated on 10/Feb/18

$$Quitegood.$$

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