int frac{x - 2}{x(2log x - x)} dx = | vedclass

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Question

(B) Let $I = \int \frac{x - 2}{x(2\log x - x)} dx$. We can rewrite the integrand as: $I = \int \frac{-(2 - x)}{x(2\log x - x)} dx = - \int \frac{\frac

Vedclass · Accepted Answer

$\log \left( \frac{1}{2\log x - x} \right) + c$

Vedclass · Answer

(B) Let $I = \int \frac{x - 2}{x(2\log x - x)} dx$. We can rewrite the integrand as: $I = \int \frac{-(2 - x)}{x(2\log x - x)} dx = - \int \frac{\frac{2}{x} - 1}{2\log x - x} dx$. Let $t = 2\log x - x$. Then,differentiating with respect to $x$,we get $dt = (\frac{2}{x} - 1) dx$. Substituting these into the integral,we get: $I = - \int \frac{1}{t} dt = - \log |t| + c$. Substituting $t = 2\log x - x$ back,we get: $I = - \log |2\log x - x| + c = \log \left( \frac{1}{|2\log x - x|} \right) + c$.

$\int \frac{x - 2}{x(2\log x - x)} dx = $

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