If $f\left(x\right)$ is a polynomial satisfying $f\left(x\right).f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right),$ and $f\left(3\right)=82,$then $\int \frac{f\left(x\right)}{x^2+1}dx=$

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

$The polynomial function satisfying$

$above equation is f\left(x\right)=1\pm x^n$

$$\begin{gathered} Given that f\left(3\right)=82 \\ =1+81 \\ f\left(3\right)=1+3^4 \\ Replace 3 by x \\ f\left(x\right)=1+x^4 \\ now \int \frac{f\left(x\right)}{x^2+1}dx=\int \frac{x^4}{x^2+1}dx \\ = \int \frac{\left(x^4-1\right)+2}{x^2+1}dx \\ =\int \left[\frac{\left(x^2+1\right)\left(x^2-1\right)}{x^2+1}+\frac{2}{x^2+1}\right]dx \\ =\int \left(x^2-1+\frac{2}{x^2+1}\right) \\ =\frac{x^3}{3}-x+2{\tan }^{-1}x+C \end{gathered}$$