If \(\displaystyle 2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1\), then \(\int f(x)\,dx\) is: (Here C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{3}\left(\frac{2}{3}x^3 + \frac{1}{x} + x\right) + C\)
Replacing x by \(\frac{1}{x}\) in the given relation gives a second equation:
\(\displaystyle 2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1 \quad \text{(i)}\)
\(\displaystyle f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x^2} + 1 \quad \text{(ii)}\)
2 × (i) − (ii) removes \(f\left(\frac{1}{x}\right)\):
\(\displaystyle 3f(x) = 2x^2 + 2 - \frac{1}{x^2} - 1\)
\(\displaystyle f(x) = \frac{1}{3}\left(2x^2 + 1 - \frac{1}{x^2}\right)\)
Integrating term by term:
\(\displaystyle \int f(x)\,dx = \frac{1}{3}\left(\frac{2}{3}x^3 + x + \frac{1}{x}\right) + C\)