int_a^b frac{log x}{x} , dx = | vedclass

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

Question

(C) Let $I = \int_a^b \frac{\log x}{x} \, dx$. Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$. When $x = a$,$u = \log a$. When $x = b$,$u = \lo

Vedclass · Accepted Answer

$\frac{1}{2} \log (ab) \log \left( \frac{b}{a} \right)$

Vedclass · Answer

(C) Let $I = \int_a^b \frac{\log x}{x} \, dx$. Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$. When $x = a$,$u = \log a$. When $x = b$,$u = \log b$. Therefore,$I = \int_{\log a}^{\log b} u \, du$. $I = \left[ \frac{u^2}{2} \right]_{\log a}^{\log b} = \frac{1}{2} [(\log b)^2 - (\log a)^2]$. Using the identity $x^2 - y^2 = (x + y)(x - y)$,we get: $I = \frac{1}{2} [(\log b + \log a)(\log b - \log a)]$. Since $\log b + \log a = \log (ab)$ and $\log b - \log a = \log \left( \frac{b}{a} \right)$,$I = \frac{1}{2} \log (ab) \log \left( \frac{b}{a} \right)$.

$\int_a^b \frac{\log x}{x} \, dx = $

Similar Questions

The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

The solution for $x$ of the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$ is

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module