int frac{x^5+x}{x^8+1} dx = | vedclass

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Question

(A) Let $I = \int \frac{x^5+x}{x^8+1} dx$. Divide numerator and denominator by $x^6$: $I = \int \frac{\frac{1}{x} + \frac{1}{x^5}}{x^2 + \frac{1}{x^6}

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} 	an ^{-1}\left(\frac{x^4-1}{\sqrt{2} x^2}ight)+c$

Vedclass · Answer

(A) Let $I = \int \frac{x^5+x}{x^8+1} dx$. Divide numerator and denominator by $x^6$: $I = \int \frac{\frac{1}{x} + \frac{1}{x^5}}{x^2 + \frac{1}{x^6}} dx$. This approach is complex. Alternative: Let $x^2 = t$, then $2x dx = dt$. $I = \frac{1}{2} \int \frac{t^2+1}{t^4+1} dt$. Divide numerator and denominator by $t^2$: $I = \frac{1}{2} \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$. Let $t - \frac{1}{t} = u$, then $(1 + \frac{1}{t^2}) dt = du$. Also, $t^2 + \frac{1}{t^2} = (t - \frac{1}{t})^2 + 2 = u^2 + 2$. $I = \frac{1}{2} \int \frac{du}{u^2 + 2} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} 	an^{-1} \left(\frac{u}{\sqrt{2}}ight) + c$. Substituting $u = t - \frac{1}{t} = \frac{t^2-1}{t} = \frac{x^4-1}{x^2}$: $I = \frac{1}{2\sqrt{2}} 	an^{-1} \left(\frac{x^4-1}{\sqrt{2}x^2}ight) + c$.

$\int \frac{x^5+x}{x^8+1} dx =$

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