int frac{1}{x} log x , dx is equal to | vedclass

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Question

(B) Let $I = \int \frac{1}{x} \log x \, dx$. Substitute $\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} \, dx = dt$.

Vedclass · Accepted Answer

$\frac{1}{2} (\log x)^2 + c$

Vedclass · Answer

(B) Let $I = \int \frac{1}{x} \log x \, dx$. Substitute $\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} \, dx = dt$. Substituting these into the integral,we get $I = \int t \, dt$. Integrating with respect to $t$,we get $I = \frac{t^2}{2} + c$. Substituting back $t = \log x$,we get $I = \frac{(\log x)^2}{2} + c$.

$\int \frac{1}{x} \log x \, dx$ is equal to

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