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) ધારો કે $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.$u = 1 + \log x$ આદેશ લેતા,$du = \frac{1}{x} dx$ મળે.જ્યારે $x = 1$,ત્યારે $u = 1 + \log 1 =

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) ધારો કે $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.$u = 1 + \log x$ આદેશ લેતા,$du = \frac{1}{x} dx$ મળે.જ્યારે $x = 1$,ત્યારે $u = 1 + \log 1 = 1 + 0 = 1$.જ્યારે $x = e$,ત્યારે $u = 1 + \log e = 1 + 1 = 2$.તેથી,$I = \int_{1}^{2} \frac{du}{u^{2}} = \int_{1}^{2} u^{-2} du$.$I = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$.$I = -\left( \frac{1}{2} - \frac{1}{1} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

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

Similar Questions

સંકલન $\int_{\pi /6}^{\pi /3} {\sec ^{2/3} x \, \csc ^{4/3} x \, dx}$ ની કિંમત શોધો.

$\int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx = $ . . . . . . .

$\int_1^2 x \sqrt{4-x^2} \, dx =$

$\int_{2}^{3} \frac{x}{x^{2}-1} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module