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x-4-1-x-2-x-4-x-2-1-dx-




Question Number 206537 by necx122 last updated on 18/Apr/24
∫((x^4 −1)/(x^2 (√(x^4 +x^2 +1))))dx
$$\int\frac{{x}^{\mathrm{4}} −\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}}{dx} \\ $$
Answered by Berbere last updated on 18/Apr/24
=∫((x(4x^3 +2x)−2x^4 −2x^2 −2)/(2x^2 (√(x^4 +x^2 +1))))  =∫((x.((√(x^4 +x^2 +1)))′)/x^2 )−((1.(√(x^4 +x^2 +1)))/x^2 )dx  =∫d(((√(x^4 +x^2 +1))/x))=((√(x^4 +x^2 +1))/x)+a;a∈R
$$=\int\frac{{x}\left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{2}{x}\right)−\mathrm{2}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\int\frac{{x}.\left(\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\right)'}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}.\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}}{{x}^{\mathrm{2}} }{dx} \\ $$$$=\int{d}\left(\frac{\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}}{{x}}\right)=\frac{\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}}{{x}}+{a};{a}\in\mathbb{R} \\ $$$$\: \\ $$

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