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

∫x^6 −1/x^2 −1dx

$$\int{x}^{\mathrm{6}} −\mathrm{1}/{x}^{\mathrm{2}} −\mathrm{1}{dx} \\ $$

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

∫ ((x^6  − 1)/(x^2  − 1)) dx  = ∫ (((x^2 )^3  − 1^3 )/(x^2  − 1)) dx  Simplify the numerator using the identity:   x^3  − y^3  = (x − y)(x^2  + xy + y^2 )  = ∫ (((x^2  − 1)[(x^2 )^2  + x^2  + 1^2 ])/(x^2  − 1)) dx  = ∫ (((x^2  − 1)(x^4  + x^2  + 1))/(x^2  − 1)) dx  = ∫ (x^4  + x^2  + 1) dx  = (x^5 /5) + (x^3 /3) + x + C  =  (1/5) x^5   +  (1/3) x^3  +  x  +  C

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

Commented by ajfour last updated on 10/Feb/18

Quite good.

$${Quite}\:{good}. \\ $$

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