int frac{f(x) g^{prime}(x)-f^{prime}(x) g(x)} | vedclass

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(B) Let $I = \int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} [\log g(x)-\log f(x)] \, dx$. We know that $\frac{d}{dx} \left( \log \frac{g

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$\frac{1}{2}\left[\log \frac{g(x)}{f(x)}\right]^2+C$

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(B) Let $I = \int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} [\log g(x)-\log f(x)] \, dx$. We know that $\frac{d}{dx} \left( \log \frac{g(x)}{f(x)} \right) = \frac{d}{dx} (\log g(x) - \log f(x)) = \frac{g^{\prime}(x)}{g(x)} - \frac{f^{\prime}(x)}{f(x)} = \frac{f(x) g^{\prime}(x) - f^{\prime}(x) g(x)}{f(x) g(x)}$. Thus,the integral becomes $I = \int \log \left[\frac{g(x)}{f(x)}\right] \cdot d\left( \log \frac{g(x)}{f(x)} \right)$. Let $t = \log \left[\frac{g(x)}{f(x)}\right]$,then $dt = d\left( \log \frac{g(x)}{f(x)} \right)$. Substituting this into the integral,we get $I = \int t \, dt = \frac{t^2}{2} + C$. Substituting back the value of $t$,we get $I = \frac{1}{2} \left[ \log \frac{g(x)}{f(x)} \right]^2 + C$.

$\int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} \times [\log g(x)-\log f(x)] \, dx$ is equal to

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