If frac{d}{d x}left(frac{x cdot 2^x-x}{1-cos | vedclass

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(B) Let $y = \frac{x \cdot 2^x-x}{1-\cos x}$. Then the given equation is $\frac{dy}{dx} = y(f(x) + \log 2)$.This implies $\frac{1}{y} \frac{dy}{dx} =

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$\frac{1}{x}+\frac{\log 2}{2^x-1}-\frac{\sin x}{1-\cos x}$

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(B) Let $y = \frac{x \cdot 2^x-x}{1-\cos x}$. Then the given equation is $\frac{dy}{dx} = y(f(x) + \log 2)$.This implies $\frac{1}{y} \frac{dy}{dx} = f(x) + \log 2$, or $\frac{d}{dx}(\ln |y|) = f(x) + \log 2$.Thus, $f(x) = \frac{d}{dx}(\ln |y|) - \log 2$.Since $y = \frac{x(2^x-1)}{1-\cos x}$, we have $\ln |y| = \ln |x| + \ln |2^x-1| - \ln |1-\cos x|$.Differentiating with respect to $x$:$\frac{d}{dx}(\ln |y|) = \frac{1}{x} + \frac{1}{2^x-1} \cdot (2^x \log 2) - \frac{\sin x}{1-\cos x}$.Substituting this into the expression for $f(x)$:$f(x) = \frac{1}{x} + \frac{2^x \log 2}{2^x-1} - \frac{\sin x}{1-\cos x} - \log 2$.$f(x) = \frac{1}{x} + \frac{2^x \log 2 - (2^x-1) \log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.$f(x) = \frac{1}{x} + \frac{2^x \log 2 - 2^x \log 2 + \log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.$f(x) = \frac{1}{x} + \frac{\log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.

If $\frac{d}{d x}\left(\frac{x \cdot 2^x-x}{1-\cos x}\right)=\left(\frac{x \cdot 2^x-x}{1-\cos x}\right)(f(x)+\log 2)$, then $f(x)=$

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