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

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Question

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

Vedclass · Accepted Answer

$\log |1 + \log x|$

Vedclass · Answer

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

$\int \frac{1}{x+x \log x} \, dx = $ . . . . . . $+ C$.

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