If $f(x)=\frac{x-1}{x+1}, f^2(x)=f(f(x)), ......, f^{k+1}(x)=f\left(f^k(x)\right), k=1,2,3, ...$. and $g(x)=f^{1998}(x)$, then $\int\limits_{1 / e}^1 g(x) d x$ is equal to

-1

We have,

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

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

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

∴  $g(x)=f^{1998}(x)=f^2 of^{1996}(x)$

$\Rightarrow g(x)=f^2\left(f^{1996}(x)\right)$

$\Rightarrow g(x)=f^2(x)$         $\left[∵ f^{1996}(x)=\left(\begin{array}{c}f^4 o f^4 o f^4 o ... of^4 \\ 499 \text { times }\end{array}\right)(x)=x\right]$

$\Rightarrow g(x)=-\frac{1}{x}$

∴   $\int\limits_{1 / e}^1 g(x) d x=\int\limits_{1 / e}^1-\frac{1}{x} d x=-\left[\log _e x\right]_{1 / e}^1$

$\Rightarrow \int\limits_{1 / e}^1 g(x) d x=-\left[\log _e 1-\log _e 1 / e\right]=-[0+1]=-1$