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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(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

Similar Questions

The value of $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$ is equal to (where $C$ is an arbitrary constant.)

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k$ where $k$ is a constant of integration,then $A+B+C$ equals

$\int \frac{x+1}{\sqrt{x^2+x+1}} d x=$

If $\int \frac{dx}{(x^2 - 2x + 10)^2} = A \left( \tan^{-1} \left( \frac{x - 1}{3} \right) + \frac{f(x)}{x^2 - 2x + 10} \right) + C$,where $C$ is a constant of integration,then:

If $\int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x (1 + \cos \frac{1}{\beta} x)} \right| + C$,then $\beta - \alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module