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Question Number 76468 by kaivan.ahmadi last updated on 27/Dec/19

∫((1−x^2 −8x+1)/(x^4 −x^3 −x+1))dx

$$\int\frac{\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}}{dx} \\ $$

Commented by mr W last updated on 27/Dec/19

why did you write 1−x^2 −8x+1 instead of 2−x^2 −8x ?

$${why}\:{did}\:{you}\:{write}\:\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1} \\ $$$${instead}\:{of}\:\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{8}{x}\:? \\ $$

Commented by kaivan.ahmadi last updated on 27/Dec/19

i dont know.it is a practice that i saw it.

$${i}\:{dont}\:{know}.{it}\:{is}\:{a}\:{practice}\:{that}\:{i}\:{saw}\:{it}. \\ $$

Answered by MJS last updated on 27/Dec/19

just decompose ∫((−x^2 −8x+2)/(x^4 −x^3 −x+1))dx=−∫((x^2 +8x−2)/((x−1)^2 (x^2 +x+1)))dx= =∫((x+((10)/3))/(x^2 +x+1))dx−(7/3)∫(dx/((x−1)^2 ))−∫(dx/(x−1))= =((17)/6)∫(dx/(x^2 +x+1))+(1/2)∫((2x+1)/(x^2 +x+1))dx−(7/3)∫(dx/((x−1)^2 ))−∫(dx/(x−1))= =((17(√3))/9)arctan (((√3)(2x+1))/3)+(1/2)ln (x^2 +x+1) +(7/(3(x−1)))−ln ∣x−1∣ = =((17(√3))/9)arctan (((√3)(2x+1))/3) +(1/2)ln ((x^2 +x+1)/((x−1)^2 )) +(7/(3(x−1))) +C

$$\mathrm{just}\:\mathrm{decompose} \\ $$$$\int\frac{−{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{2}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}}{dx}=−\int\frac{{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\int\frac{{x}+\frac{\mathrm{10}}{\mathrm{3}}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{7}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\int\frac{{dx}}{{x}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{17}}{\mathrm{6}}\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{7}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\int\frac{{dx}}{{x}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{17}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:+\frac{\mathrm{7}}{\mathrm{3}\left({x}−\mathrm{1}\right)}−\mathrm{ln}\:\mid{x}−\mathrm{1}\mid\:= \\ $$$$=\frac{\mathrm{17}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{7}}{\mathrm{3}\left({x}−\mathrm{1}\right)}\:+{C} \\ $$

Commented by kaivan.ahmadi last updated on 29/Dec/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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