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$\int \frac{x^8+4}{x^4-2 x^2+2} d x=A x^5+B x^3+C x+k$, then $5 A+3 B+C=$
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$5$
$\begin{aligned} & \int \frac{x^8+4}{x^4-2 x^2+2} d x=\int \frac{x^8+4 x^4+4-4 x^4}{x^4-2 x^2+2} d x \\ & =\int \frac{\left(x^4+2\right)^2-\left(2 x^2\right)^2}{x^4-2 x^2+2} d x=\int x^4+2+2 x^2 d x \\ & =\frac{x^5}{5}+\frac{2 x^3}{3}+2 x+k \\ \Rightarrow & A=\frac{1}{5}, B=\frac{2}{3}, C=2 \\ & 5 A+3 B+C=5\end{aligned}$
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