To find the value of int frac{1 + log x}{x} d | vedclass

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(B) Let $1 + \log x = t$. Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(1 + \log x) = \frac{dt}{dx}$. This implies $\frac{1}{x}

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$1 + \log x = t$

Vedclass · Answer

(B) Let $1 + \log x = t$. Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(1 + \log x) = \frac{dt}{dx}$. This implies $\frac{1}{x} dx = dt$. Substituting these into the integral,we get $\int (1 + \log x) \cdot \frac{1}{x} dx = \int t dt$. Integrating with respect to $t$,we get $\frac{t^2}{2} + C$. Substituting back $t = 1 + \log x$,the final result is $\frac{(1 + \log x)^2}{2} + C$.

To find the value of $\int \frac{1 + \log x}{x} dx$,the proper substitution is

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