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Question Number 194456 by horsebrand11 last updated on 07/Jul/23

Answered by cortano12 last updated on 07/Jul/23

$$\underbrace{⋚}$$

Answered by Frix last updated on 08/Jul/23

$$\frac{x^8-1}{\sqrt{x^{14}+x^6}}=\frac{x^8-1}{x^3\sqrt{x^8+1}}$$$$\mathrm{Without}\mathrm{integration},\mathrm{just}\mathrm{following}\mathrm{an}\mathrm{idea}:$$$$\frac{d\left[\frac{\sqrt{x^8+1}}{x^n}\right]}{dx}=\frac{(4-n)x^8-n}{x^{n+1}\sqrt{x^8+1}}$$$$\Rightarrow n=2$$$$\frac{d\left[\frac{\sqrt{x^8+1}}{x^2}\right]}{dx}=\frac{2x^8-2}{x^3\sqrt{x^8+1}}$$$$\Rightarrow$$$$\int \frac{x^8-1}{x^3\sqrt{x^8+1}}dx=\frac{\sqrt{x^8+1}}{2x^2}+C$$

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