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

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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 $C$.

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

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