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

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

Question

(A) Let $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.Substitute $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.When $x = 1$,$u = 1 + \log 1 = 1 + 0 = 1

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.Substitute $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.When $x = 1$,$u = 1 + \log 1 = 1 + 0 = 1$.When $x = e$,$u = 1 + \log e = 1 + 1 = 2$.Thus,$I = \int_{1}^{2} \frac{du}{u^{2}} = \int_{1}^{2} u^{-2} du$.$I = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$.$I = -\left( \frac{1}{2} - \frac{1}{1} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

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

Similar Questions

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$

$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to

$\int_0^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx =$

$\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x=$ . . . . . . .

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module