If $2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1$, then $\int f(x) dx$ is: (Here C is an arbitrary constant)

Given: $2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1$ ... (Equation 1)

Replace $x$ with $\frac{1}{x}$ in the original equation:

$2f\left(\frac{1}{x}\right) + f\left(\frac{1}{\frac{1}{x}}\right) = \left(\frac{1}{x}\right)^2 + 1$

$2f\left(\frac{1}{x}\right) + f(x) = \frac{1}{x^2} + 1$ ... (Equation 2)

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From the two equations:

Equation 1: $2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1$

Equation 2: $f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x^2} + 1$

Multiply Equation 1 by 2:

$4f(x) + 2f\left(\frac{1}{x}\right) = 2x^2 + 2$

Subtract Equation 2:

$4f(x) + 2f\left(\frac{1}{x}\right) - f(x) - 2f\left(\frac{1}{x}\right) = 2x^2 + 2 - \frac{1}{x^2} - 1$

$3f(x) = 2x^2 + 1 - \frac{1}{x^2}$

$f(x) = \frac{1}{3}\left(2x^2 - \frac{1}{x^2} + 1\right)$

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$\int f(x) dx = \int \frac{1}{3}\left(2x^2 - \frac{1}{x^2} + 1\right) dx$

$= \frac{1}{3} \int \left(2x^2 - \frac{1}{x^2} + 1\right) dx$

$= \frac{1}{3} \int \left(2x^2 - x^{-2} + 1\right) dx$

$= \frac{1}{3}\left(\frac{2x^3}{3} - \frac{x^{-1}}{-1} + x\right) + C$

$= \frac{1}{3}\left(\frac{2x^3}{3} + \frac{1}{x} + x\right) + C$

Therefore, $\int f(x) dx = \frac{1}{3}\left(\frac{2x^3}{3} + \frac{1}{x} + x\right) + C$