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

If $\int \frac{dx}{x^{7/2}(x^4+1)^{3/8}} = m \left(\frac{x^4+1}{x^4}\right)^n + c$, where $c$ is a constant of integration, then the value of $n$ is...

For $x>0$, evaluate the integral: $\int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$.

Integrate the function: $\sqrt{x^{2}+4x+1}$

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module