If int frac{dx}{x(log x-2)(log x-3)}=I+C, the | vedclass

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(B) Let $I = \int \frac{dx}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$, so $dt = \frac{dx}{x}$.Then the integral becomes $I = \int \frac{dt}{(t-2

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$\log \left|\frac{\log x-3}{\log x-2}\right|$

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(B) Let $I = \int \frac{dx}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$, so $dt = \frac{dx}{x}$.Then the integral becomes $I = \int \frac{dt}{(t-2)(t-3)}$.Using partial fractions: $\frac{1}{(t-2)(t-3)} = \frac{A}{t-2} + \frac{B}{t-3}$.$1 = A(t-3) + B(t-2)$.For $t=2$, $A = -1$. For $t=3$, $B = 1$.So, $I = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) dt$.$I = \log |t-3| - \log |t-2| + C$.$I = \log \left| \frac{t-3}{t-2} \right| + C$.Substituting $t = \log x$ back, we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right| + C$.

If $\int \frac{dx}{x(\log x-2)(\log x-3)}=I+C$, then $I$ is equal to

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