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

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

The correct answer is Option (1) → $\frac{1}{3}\left(\frac{2}{3}x^3+\frac{1}{x}+x\right)+C$

Given equation:

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

Replace \(x\) by \(\frac{1}{x}\):

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

System of equations:

\[ \begin{cases} 2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1 \\ f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x^2} + 1 \end{cases} \]

Multiply first equation by 2:

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

Subtract second equation:

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

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

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

Integrate \(f(x)\):

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

\[ = \frac{1}{3} \left( \frac{2x^3}{3} + x \right) - \frac{1}{3} \left( -x^{-1} \right) + C = \frac{2x^3}{9} + \frac{x}{3} + \frac{1}{3x} + C \]

\({\int f(x) \, dx = \frac{2x^3}{9} + \frac{x}{3} + \frac{1}{3x} + C}\)