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

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(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.The integral becomes $I_1 = \int \frac{d t

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

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(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.The integral becomes $I_1 = \int \frac{d t}{(t-2)(t-3)}$.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 $t=3$,$B=1$. For $t=2$,$A=-1$.So,$I_1 = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) d t$.$I_1 = \log |t-3| - \log |t-2| + C = \log \left| \frac{t-3}{t-2} \right| + C$.Substituting $t = \log x$ back,we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right|$.

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

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