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6x-4-4-x-4-2-dx-




Question Number 85441 by jagoll last updated on 22/Mar/20
∫ ((6x^4 −4)/( (√(x^4 −2)))) dx = ?
$$\int\:\frac{\mathrm{6x}^{\mathrm{4}} −\mathrm{4}}{\:\sqrt{\mathrm{x}^{\mathrm{4}} −\mathrm{2}}}\:\mathrm{dx}\:=\:? \\ $$
Answered by john santu last updated on 22/Mar/20
I = ∫ ((4x^4 )/( (√(x^4 −2)))) dx = 4∫ x ((x^3 /( (√(x^4 −2))))) dx   = 4 ∫ (1/4)x ((d(x^4 −2))/( (√(x^4 −2))))  = 4 ((1/4)x(2(√(x^4 −2)))−∫ (1/2)(√(x^4 −2))) + c  = 2x(√(x^4 −2)) −2∫(√(x^4 −2)) dx  J = ∫ ((2x^4 −4)/( (√(x^4 −2)))) dx = 2∫ ((x^4 −2)/( (√(x^4 −2)))) dx  J = 2∫ (√(x^4 −2)) dx  the answer   I + J = 2x(√(x^4 −2)) −2∫(√(x^4 −2 ))dx +   2 ∫ (√(x^4 −2)) dx = 2x(√(x^4 −2)) + c . ∴
$${I}\:=\:\int\:\frac{\mathrm{4}{x}^{\mathrm{4}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}}\:{dx}\:=\:\mathrm{4}\int\:{x}\:\left(\frac{{x}^{\mathrm{3}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}}\right)\:{dx}\: \\ $$$$=\:\mathrm{4}\:\int\:\frac{\mathrm{1}}{\mathrm{4}}{x}\:\frac{{d}\left({x}^{\mathrm{4}} −\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}} \\ $$$$=\:\mathrm{4}\:\left(\frac{\mathrm{1}}{\mathrm{4}}{x}\left(\mathrm{2}\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\right)−\int\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\right)\:+\:{c} \\ $$$$=\:\mathrm{2}{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:−\mathrm{2}\int\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:{dx} \\ $$$${J}\:=\:\int\:\frac{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{4}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}}\:{dx}\:=\:\mathrm{2}\int\:\frac{{x}^{\mathrm{4}} −\mathrm{2}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}}\:{dx} \\ $$$${J}\:=\:\mathrm{2}\int\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:{dx} \\ $$$${the}\:{answer}\: \\ $$$${I}\:+\:{J}\:=\:\mathrm{2}{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:−\mathrm{2}\int\sqrt{{x}^{\mathrm{4}} −\mathrm{2}\:}{dx}\:+\: \\ $$$$\mathrm{2}\:\int\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:{dx}\:=\:\mathrm{2}{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}\:+\:{c}\:.\:\therefore \\ $$
Commented by jagoll last updated on 22/Mar/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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