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calculate-x-3-4x-5-x-2-x-1-dx-




Question Number 73477 by abdomathmax last updated on 13/Nov/19
calculate ∫   ((x^3 −4x+5)/(x^2 −x +1))dx
$${calculate}\:\int\:\:\:\frac{{x}^{\mathrm{3}} −\mathrm{4}{x}+\mathrm{5}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}{dx} \\ $$
Answered by MJS last updated on 13/Nov/19
∫((x^3 −4x+5)/(x^2 −x+1))dx=  =∫xdx+∫dx−4∫((x−1)/(x^2 −x+1))dx  ∫((x−1)/(x^2 −x+1))dx=(1/2)∫((2x−1)/(x^2 −x+1))dx−(1/2)∫(dx/(x^2 −x+1))=  =(1/2)ln (x^2 −x+1) −((√3)/3)arctan (((2x−1)(√3))/3)  ∫((x^3 −4x+5)/(x^2 −x+1))dx=  =(1/2)x^2 +x−2ln (x^2 −x+1) +((4(√3))/3)arctan (((2x−1)(√3))/3) +C
$$\int\frac{{x}^{\mathrm{3}} −\mathrm{4}{x}+\mathrm{5}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=\int{xdx}+\int{dx}−\mathrm{4}\int\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$\int\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\int\frac{{x}^{\mathrm{3}} −\mathrm{4}{x}+\mathrm{5}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{x}−\mathrm{2ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\:+{C} \\ $$

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