int frac{1}{x+x log x} d x= . . . . . . . | vedclass

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

Question

(C) To evaluate the integral $I = \int \frac{1}{x+x \log x} dx$,we first factor out $x$ from the denominator:$I = \int \frac{1}{x(1+\log x)} dx$Let $u

Vedclass · Accepted Answer

$\log |1+\log x| + C$

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{1}{x+x \log x} dx$,we first factor out $x$ from the denominator:$I = \int \frac{1}{x(1+\log x)} dx$Let $u = 1+\log x$. Then,the derivative is $du = \frac{1}{x} dx$.Substituting these into the integral,we get:$I = \int \frac{1}{u} du$$I = \log |u| + C$Substituting back $u = 1+\log x$,we obtain:$I = \log |1+\log x| + C$Thus,the correct option is $C$.

$\int \frac{1}{x+x \log x} d x=$ . . . . . . .

Similar Questions

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = $

The value of $\int \frac{e^x}{e^x + 1} \,dx$ is

$\int(x+2) \sqrt{x+3} \, dx =$

$\int \frac{dx}{x\sqrt{1 - (\log x)^2}} = $

$\int {{x^3}\sqrt {3 + 5{x^4}} } \;dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module