If int_2^{e}left[frac{1}{log x}-frac{1}{(log | vedclass

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

Question

(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$

Vedclass · Accepted Answer

$a=e, b=-2$

Vedclass · Answer

(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$. When $x = 2$,$t = \log 2$. When $x = e$,$t = 1$. Substituting these into the integral: $I = \int_{\log 2}^{1} \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt$. We know that $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$. Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$. Therefore,$I = \left[ e^t \cdot \frac{1}{t} \right]_{\log 2}^{1}$. $I = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log 2} \cdot \frac{1}{\log 2} \right)$. $I = e - \frac{2}{\log 2}$. Comparing this with $a + \frac{b}{\log 2}$,we get $a = e$ and $b = -2$.

If $\int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx = a+\frac{b}{\log 2}$,then:

Similar Questions

$\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) dx$ is equal to

$\int \frac{(x-3) e^x}{(x-1)^3} d x=$ . . . . . . $+C$.

$\int {\frac{{x - 1}}{{{{(x + 1)}^3}}}{e^x}\,dx = } $

If $\int e^{2x} f^{\prime}(x) dx = g(x)$, then $\int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx =$

If $\int e^{\sin x} \cdot \left[ \frac{x \cos^3 x - \sin x}{\cos^2 x} \right] dx = e^{\sin x} f(x) + c$, where $c$ is the constant of integration, then $f(x)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module