int frac{dx}{x + x log x} = | vedclass

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Question

(A) We are given the integral $I = \int \frac{dx}{x + x \log x}$.First,factor out $x$ from the denominator:$I = \int \frac{dx}{x(1 + \log x)}$.Now,use

Vedclass · Accepted Answer

$\log(1 + \log x) + C$

Vedclass · Answer

(A) We are given the integral $I = \int \frac{dx}{x + x \log x}$.First,factor out $x$ from the denominator:$I = \int \frac{dx}{x(1 + \log x)}$.Now,use the method of substitution. Let $t = 1 + \log x$.Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $\frac{dx}{x} = dt$.Substituting these into the integral,we get:$I = \int \frac{dt}{t}$.The integral of $\frac{1}{t}$ with respect to $t$ is $\log|t| + C$.Substituting back $t = 1 + \log x$,we get:$I = \log|1 + \log x| + C$.

$\int \frac{dx}{x + x \log x} = $

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