Question

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

• A. $x^3-x+2{\tan }^{-1}x+c$
• B. $\frac{1}{3}{x}^3-x+{\tan }^{-1}x+c$
• C. $\frac{x^3}{3}-x+2{\tan }^{-1}x+c$
• D. $\frac{1}{3}{x}^3+x+2{\tan }^{-1}x+c$

Answer 1

Answer: (C)

$\frac{x^3}{3}-x+2{\tan }^{-1}x+c$

Given : $f\left(3\right)=82=81+1$, from the given equation we can find $f\left(\frac{1}{x}\right)$

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

$⟹f\left(x\right)=\left\{f\right(x)-1\}f\left(\frac{1}{x}\right)$

$⟹f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{f\left(x\right)-1}$

$\therefore$ we get $f\left(\frac{1}{3}\right)=\frac{82}{81}=1+\frac{1}{81}$

$\therefore f\left(3\right)=3^4+1$ and $f\left(\frac{1}{3}\right)=1+\frac{1}{3^4}$.

Therefore we can observe that $f\left(x\right)=x^4+1$

$⟹\int \frac{f\left(x\right)}{x^2+1}dx=\int \frac{x^4+1}{x^2+1}dx=\int \frac{x^4-1}{x^2+1}dx+\int \frac{2}{x^2+1}dx=(\int (x^2-1\left)dx\right)+2{\tan }^{-1}x=\frac{x^3}{3}-x+2{\tan }^{-1}x+c$

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