int_{1}^{x} frac{log(x^2)}{x} , dx = | vedclass

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

Question

(A) Given integral: $I = \int_{1}^{x} \frac{\log(x^2)}{x} \, dx$ Using the property $\log(a^b) = b \log a$,we have: $I = \int_{1}^{x} \frac{2 \log x}{

Vedclass · Accepted Answer

$(\log x)^2$

Vedclass · Answer

(A) Given integral: $I = \int_{1}^{x} \frac{\log(x^2)}{x} \, dx$ Using the property $\log(a^b) = b \log a$,we have: $I = \int_{1}^{x} \frac{2 \log x}{x} \, dx$ Let $t = \log x$,then $dt = \frac{1}{x} \, dx$. When $x = 1$,$t = \log(1) = 0$. When $x = x$,$t = \log x$. Substituting these into the integral: $I = \int_{0}^{\log x} 2t \, dt$ $I = 2 \left[ \frac{t^2}{2} \right]_{0}^{\log x}$ $I = [t^2]_{0}^{\log x}$ $I = (\log x)^2 - 0^2 = (\log x)^2$ Thus,the correct option is $A$.

$\int_{1}^{x} \frac{\log(x^2)}{x} \, dx = $

Similar Questions

If $\int \limits_{\frac{1}{3}}^3 |\log_e x| dx = \frac{m}{n} \log_e \left(\frac{n^2}{e}\right)$,where $m$ and $n$ are coprime natural numbers,then $m^2 + n^2 - 5$ is equal to $............$.

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

$\int_{0}^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) d(x - [x])$ is : (where $[.]$ denotes the greatest integer function)

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

The value of $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module