If f_n(x) = log log log ldots log x (where lo | vedclass

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

Question

(A) Given $f_n(x) = \log \log \ldots \log x$ ($n$ times). Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$. Let $t = f_n(x) = \log(f_{n-1}(x))$

Vedclass · Accepted Answer

$f_{n+1}(x) + c$

Vedclass · Answer

(A) Given $f_n(x) = \log \log \ldots \log x$ ($n$ times). Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$. Let $t = f_n(x) = \log(f_{n-1}(x))$. Then, $\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$. Thus, $dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$. Substituting this into the integral: $I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)} = \int \frac{dt}{t}$. $I = \log(t) + c = \log(f_n(x)) + c$. Since $f_{n+1}(x) = \log(f_n(x))$, we have $I = f_{n+1}(x) + c$.

If $f_n(x) = \log \log \log \ldots \log x$ (where $\log$ is repeated $n$ times), then $\int (x f_1(x) f_2(x) \ldots f_n(x))^{-1} dx$ is equal to

Similar Questions

$\int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx = $

$\int \frac{2+\cos \frac{x}{2}}{x+\sin \frac{x}{2}} \,d x=$

Let $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. If $I(0)=3$,then $I\left(\frac{\pi}{12}\right)$ is equal to :

$\int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)} = $

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$ is equal to-

Vedclass Test Series

Exam Paper Generator

Online Exam Module