If the graph of the anti-derivative g(x) of f | vedclass

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(C) Given: $f(x) = \log(\log x) + (\log x)^{-2}$.Anti-derivative $g(x) = \int (\log(\log x) + (\log x)^{-2}) dx$.Let $t = \log x$,then $x = e^t$ and $

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$7$

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(C) Given: $f(x) = \log(\log x) + (\log x)^{-2}$.Anti-derivative $g(x) = \int (\log(\log x) + (\log x)^{-2}) dx$.Let $t = \log x$,then $x = e^t$ and $dx = e^t dt$.Substituting these into the integral:$g(x) = \int e^t (\log t + t^{-2}) dt$.We can rewrite the integrand as $e^t (\log t + t^{-1} - t^{-1} + t^{-2}) = e^t (\log t + t^{-1}) + e^t (-t^{-1} + t^{-2})$.Using the formula $\int e^t (h(t) + h'(t)) dt = e^t h(t) + C$,where $h(t) = \log t$,we get:$g(x) = e^t \log t - e^t t^{-1} + C = e^t (\log t - t^{-1}) + C$.Substituting $t = \log x$ back:$g(x) = x (\log(\log x) - (\log x)^{-1}) + C$.Since the graph passes through $(e, 2023 - e)$:$2023 - e = e (\log(\log e) - (\log e)^{-1}) + C$.$2023 - e = e (\log(1) - 1) + C$.$2023 - e = e (0 - 1) + C = -e + C$.Thus,$C = 2023$.The term independent of $x$ is $k = 2023$.The sum of the digits of $k$ is $2 + 0 + 2 + 3 = 7$.

If the graph of the anti-derivative $g(x)$ of $f(x) = \log(\log x) + (\log x)^{-2}$ passes through $(e, 2023 - e)$ and the term independent of $x$ in $g(x)$ is $k$,then the sum of all the digits of $k$ is

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