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Question Number 97591 by  M±th+et+s last updated on 08/Jun/20

∫(x^4 /(x^3 −2x^2 −7x+4))dx

$$\int\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}}{dx} \\ $$

Answered by MJS last updated on 08/Jun/20

∫(x^4 /(x^3 −2x^2 −7x+4))dx=∫(x+2)dx+∫((11x^2 +10x−8)/(x^3 −2x^2 −7x+4))dx x^3 −2x^2 −7x+4=(x−α)(x−α)(x−γ) α=(2/3)−((10)/3)cos ((π/6)+(1/3)arcsin ((17)/(125))) β=(2/3)−((10)/3)sin ((1/3)arcsin ((17)/(125))) γ=(2/3)+((10)/3)sin ((π/3)+(1/3)arcsin ((17)/(125))) ∫((11x^2 +10x−8)/(x^3 −2x^2 −7x+4))dx= =∫((A/(x−α))+(B/(x−β))+(C/(x−γ)))dx= =Aln ∣x−α∣ +Bln ∣x−β∣ +Cln ∣x−γ∣ A=((11α^2 +10α−8)/((α−β)(α−γ))) B=((11β^2 +10β−8)/((β−α)(β−γ))) C=((11γ^2 +10γ−8)/((γ−α)(γ−β))) no easier path

$$\int\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}}{dx}=\int\left({x}+\mathrm{2}\right){dx}+\int\frac{\mathrm{11}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{8}}{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}}{dx} \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}=\left({x}−\alpha\right)\left({x}−\alpha\right)\left({x}−\gamma\right) \\ $$$$\alpha=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{10}}{\mathrm{3}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{17}}{\mathrm{125}}\right) \\ $$$$\beta=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{10}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{17}}{\mathrm{125}}\right) \\ $$$$\gamma=\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{17}}{\mathrm{125}}\right) \\ $$$$\int\frac{\mathrm{11}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{8}}{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}}{dx}= \\ $$$$=\int\left(\frac{{A}}{{x}−\alpha}+\frac{{B}}{{x}−\beta}+\frac{{C}}{{x}−\gamma}\right){dx}= \\ $$$$={A}\mathrm{ln}\:\mid{x}−\alpha\mid\:+{B}\mathrm{ln}\:\mid{x}−\beta\mid\:+{C}\mathrm{ln}\:\mid{x}−\gamma\mid \\ $$$${A}=\frac{\mathrm{11}\alpha^{\mathrm{2}} +\mathrm{10}\alpha−\mathrm{8}}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)} \\ $$$${B}=\frac{\mathrm{11}\beta^{\mathrm{2}} +\mathrm{10}\beta−\mathrm{8}}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)} \\ $$$${C}=\frac{\mathrm{11}\gamma^{\mathrm{2}} +\mathrm{10}\gamma−\mathrm{8}}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)} \\ $$$$\mathrm{no}\:\mathrm{easier}\:\mathrm{path} \\ $$

Commented by  M±th+et+s last updated on 08/Jun/20

Commented by bobhans last updated on 09/Jun/20

golden..job

$$\mathrm{golden}..\mathrm{job} \\ $$

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