$\int_1^e \frac{1 + \log x}{x} \, dx = $
- A$\frac{3}{2}$
- B$\frac{1}{2}$
- C$\frac{1}{e}$
- DNone of these
Explore More
Similar Questions
$\int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$ is equal to :
The value of $\int_0^{\pi /2} \frac{\sin x}{1 + \cos^2 x} \, dx$ is
Easy
View SolutionIf $\int\limits_0^2 375 x^5 (1 + x^2)^{-4} dx = 2^n$,then the value of $n$ is:
$\int\limits_0^1 {\frac{{{{\tan }^{ - 1}}x}}{x}\,dx} = $
The value of the integral $\int_0^{\pi /4} \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx$ equals
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo