Question Number 194456 by horsebrand11 last updated on 07/Jul/23
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Answered by cortano12 last updated on 07/Jul/23
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Answered by Frix last updated on 08/Jul/23
![((x^8 −1)/( (√(x^(14) +x^6 ))))=((x^8 −1)/(x^3 (√(x^8 +1)))) Without integration, just following an idea: ((d[((√(x^8 +1))/x^n )])/dx)=(((4−n)x^8 −n)/(x^(n+1) (√(x^8 +1)))) ⇒ n=2 ((d[((√(x^8 +1))/x^2 )])/dx)=((2x^8 −2)/(x^3 (√(x^8 +1)))) ⇒ ∫((x^8 −1)/(x^3 (√(x^8 +1))))dx=((√(x^8 +1))/(2x^2 ))+C](Q194460.png)
$$\frac{{x}^{\mathrm{8}} −\mathrm{1}}{\:\sqrt{{x}^{\mathrm{14}} +{x}^{\mathrm{6}} }}=\frac{{x}^{\mathrm{8}} −\mathrm{1}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{8}} +\mathrm{1}}} \\ $$$$\mathrm{Without}\:\mathrm{integration},\:\mathrm{just}\:\mathrm{following}\:\mathrm{an}\:\mathrm{idea}: \\ $$$$\frac{{d}\left[\frac{\sqrt{{x}^{\mathrm{8}} +\mathrm{1}}}{{x}^{{n}} }\right]}{{dx}}=\frac{\left(\mathrm{4}−{n}\right){x}^{\mathrm{8}} −{n}}{{x}^{{n}+\mathrm{1}} \sqrt{{x}^{\mathrm{8}} +\mathrm{1}}} \\ $$$$\Rightarrow\:{n}=\mathrm{2} \\ $$$$\frac{{d}\left[\frac{\sqrt{{x}^{\mathrm{8}} +\mathrm{1}}}{{x}^{\mathrm{2}} }\right]}{{dx}}=\frac{\mathrm{2}{x}^{\mathrm{8}} −\mathrm{2}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{8}} +\mathrm{1}}} \\ $$$$\Rightarrow \\ $$$$\int\frac{{x}^{\mathrm{8}} −\mathrm{1}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{8}} +\mathrm{1}}}{dx}=\frac{\sqrt{{x}^{\mathrm{8}} +\mathrm{1}}}{\mathrm{2}{x}^{\mathrm{2}} }+{C} \\ $$