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$\int \frac{5\left(x^6+1\right)}{x^2+1} \mathrm{~d} x=$ (Where $C$ is a constant of integration.)
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$x^5-\frac{5 x^3}{3}+5 x+C$
$\begin{aligned} & \int \frac{5\left(x^6+1\right)}{x^2+1} \mathrm{~d} x=\int \frac{5\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^2+1} \mathrm{~d} x \\ & =5 \int\left(x^4-x^2+1\right) \mathrm{d} x \\ & =5\left\{\frac{x^5}{5}-\frac{x^3}{3}+x\right\}+C \\ & =x^5-\frac{5}{3} x^3+5 x+C\end{aligned}$
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