$f$ the graph of the antiderivative $F (x)$ of $f (x) = log (log x) + {(log x)}^{- 2}$
passes through $(e, 1998 - e)$ then the term independent of $x$ in $F (x)$ is

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Solution

Antiderivative of $f (x) = F (x) = \int (log (log x) + {(log x)}^{- 2}) d x + C$
$= x log (log x) - \int \frac{x}{x log x} d x + \int {(log x)}^{- 2} d x + C = x log (log x) - [x {(log x)}^{- 1} + \int {(log x)}^{- 2} d x] + \int {(log x)}^{- 2} d x + C = x log (log x) - x {(log x)}^{- 1} + C$
For point $(e, 1998 - e)$ , we get
$1998 - e = e .0 - e + C \Rightarrow C = 1998$

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f the graph of the antiderivative F(x) of f(x)=log(logx)+(logx)−2 passes through (e,1998−e) then the term independent of x in F(x) is

Antiderivative of f(x)=F(x)=∫(log(logx)+(logx)−2)dx+C=xlog(logx)−∫xxlogxdx+∫(logx)−2dx+C=xlog(logx)−[x(logx)−1+∫(logx)−2dx]+∫(logx)−2dx+C=xlog(logx)−x(logx)−1+CFor point (e,1998−e), we get1998−e=e.0−e+C⇒C=1998

$f$ the graph of the antiderivative $F (x)$ of $f (x) = log (log x) + {(log x)}^{- 2}$
passes through $(e, 1998 - e)$ then the term independent of $x$ in $F (x)$ is