If int_{log _{e} 2}^{x} (e^{t}-1)^{-1} dt = l | vedclass

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

Question

(C) Given the integral $\int_{\log _{e} 2}^{x} \frac{1}{e^{t}-1} dt = \log _{e} \frac{3}{2}$.Let $u = e^{t}-1$,then $du = e^{t} dt$,which implies $dt

Vedclass · Accepted Answer

$\log _{e} 4$

Vedclass · Answer

(C) Given the integral $\int_{\log _{e} 2}^{x} \frac{1}{e^{t}-1} dt = \log _{e} \frac{3}{2}$.Let $u = e^{t}-1$,then $du = e^{t} dt$,which implies $dt = \frac{du}{u+1}$.Substituting this into the integral,we get $\int \frac{1}{u(u+1)} du = \int (\frac{1}{u} - \frac{1}{u+1}) du = \log _{e} |u| - \log _{e} |u+1| = \log _{e} |\frac{u}{u+1}|$.Substituting $u = e^{t}-1$,the integral becomes $[\log _{e} |\frac{e^{t}-1}{e^{t}}|]_{\log _{e} 2}^{x} = [\log _{e} |1-e^{-t}|]_{\log _{e} 2}^{x}$.Evaluating the limits: $\log _{e} (1-e^{-x}) - \log _{e} (1-e^{-\log _{e} 2}) = \log _{e} (1-e^{-x}) - \log _{e} (1-\frac{1}{2}) = \log _{e} (1-e^{-x}) - \log _{e} (\frac{1}{2}) = \log _{e} \frac{3}{2}$.Thus,$\log _{e} (\frac{1-e^{-x}}{1/2}) = \log _{e} \frac{3}{2}$,which implies $2(1-e^{-x}) = \frac{3}{2}$.$1-e^{-x} = \frac{3}{4} \implies e^{-x} = 1 - \frac{3}{4} = \frac{1}{4}$.Taking the natural logarithm on both sides,$-x = \log _{e} (\frac{1}{4}) = -\log _{e} 4$.Therefore,$x = \log _{e} 4$.

If $\int_{\log _{e} 2}^{x} (e^{t}-1)^{-1} dt = \log _{e} \frac{3}{2}$,then the value of $x$ is

Similar Questions

If $f$ is integrable on $[0, a]$,then the function $h$ defined on $[0, a]$ as $h(x) = \int_0^x f(t) dt$ is integrable on $[0, a]$. Which of the following functions is also integrable on $[0, a]$?

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

$\int_0^{\pi /4} (\cos x - \sin x) dx + \int_{\pi /4}^{5\pi /4} (\sin x - \cos x) dx + \int_{2\pi }^{\pi /4} (\cos x - \sin x) dx$ is equal to

The value of the integral $\int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$ is

$\int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module