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BeMath-x-5-x-x-8-1-dx-




Question Number 108169 by bemath last updated on 15/Aug/20
      ((✓BeMath✓)/(≻≺))  ∫ ((x^5 −x)/(x^8 −1)) dx ?
$$\:\:\:\:\:\:\frac{\checkmark\mathcal{B}{e}\mathcal{M}{ath}\checkmark}{\succ\prec} \\ $$$$\int\:\frac{{x}^{\mathrm{5}} −{x}}{{x}^{\mathrm{8}} −\mathrm{1}}\:{dx}\:? \\ $$
Answered by Dwaipayan Shikari last updated on 15/Aug/20
∫((x(x^4 −1))/((x^8 −1)))dx  ∫(x/(x^4 +1))dx      x^2 =u   ,2x=(du/dx)  (1/2)∫(du/(u^2 +1))=(1/2)tan^(−1) u+C=(1/2)tan^(−1) (x^2 )+C
$$\int\frac{{x}\left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\left({x}^{\mathrm{8}} −\mathrm{1}\right)}{dx} \\ $$$$\int\frac{{x}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}\:\:\:\:\:\:{x}^{\mathrm{2}} ={u}\:\:\:,\mathrm{2}{x}=\frac{{du}}{{dx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {u}+{C}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{C} \\ $$

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