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x-5x-4-x-3-5x-1-dx-




Question Number 167598 by cortano1 last updated on 20/Mar/22
      ∫ (x/(5x^4 +x^3 −5x−1)) dx=?
$$\:\:\:\:\:\:\int\:\frac{\mathrm{x}}{\mathrm{5x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} −\mathrm{5x}−\mathrm{1}}\:\mathrm{dx}=? \\ $$
Answered by MJS_new last updated on 20/Mar/22
∫(x/(5x^4 −5x+x^3 −1))dx=∫(x/((5x+1)(x^3 −1)))dx=  =∫(((25)/(126(5x+1)))+(1/(18(x−1)))−((2x+1)/(21(x^2 +x+1)))−(2/(21(x^2 +x+1))))dx=  =(5/(126))ln ∣5x+1∣ +(1/(18))ln ∣x−1∣ −(1/(21))ln (x^2 +x+1) −((4(√3))/(63))arctan (((√3)(2x+1))/3) +C
$$\int\frac{{x}}{\mathrm{5}{x}^{\mathrm{4}} −\mathrm{5}{x}+{x}^{\mathrm{3}} −\mathrm{1}}{dx}=\int\frac{{x}}{\left(\mathrm{5}{x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{dx}= \\ $$$$=\int\left(\frac{\mathrm{25}}{\mathrm{126}\left(\mathrm{5}{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{18}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{21}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{21}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}\right){dx}= \\ $$$$=\frac{\mathrm{5}}{\mathrm{126}}\mathrm{ln}\:\mid\mathrm{5}{x}+\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{18}}\mathrm{ln}\:\mid{x}−\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{21}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{63}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+{C} \\ $$

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