If int e^x cos x , dx = frac{e^x}{2}(cos x + | vedclass

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(C) Let $I = \int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx$.Substitute $t = \log \left(\frac{2x+3}{3-2x}\right)$,

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$\log f(1)$

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(C) Let $I = \int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx$.Substitute $t = \log \left(\frac{2x+3}{3-2x}\right)$, so $\frac{2x+3}{3-2x} = e^t$.Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( \frac{2x+3}{3-2x} \right) = \frac{(3-2x)(2) - (2x+3)(-2)}{(3-2x)^2} = \frac{6-4x+4x+6}{(3-2x)^2} = \frac{12}{(3-2x)^2}$.Thus, $\frac{12}{(3-2x)^2} \, dx = e^t \, dt$, which implies $\frac{1}{(3-2x)^2} \, dx = \frac{1}{12} e^t \, dt$.Substituting into the integral: $I = \int \cos(t) \cdot \frac{1}{12} e^t \, dt = \frac{1}{12} \int e^t \cos(t) \, dt$.Using the given formula $\int e^t \cos(t) \, dt = \frac{e^t}{2}(\cos t + \sin t)$, we get $I = \frac{1}{12} \cdot \frac{e^t}{2}(\cos t + \sin t) = \frac{e^t}{24}(\cos t + \sin t)$.Substituting back $t = \log \left(\frac{2x+3}{3-2x}\right)$ and $e^t = \frac{2x+3}{3-2x}$, we have $I = \frac{\frac{2x+3}{3-2x}}{24} \left[ \cos \left( \log \left( \frac{2x+3}{3-2x} \right) \right) + \sin \left( \log \left( \frac{2x+3}{3-2x} \right) \right) \right] + c$.Comparing with the given form, $f(x) = \frac{2x+3}{3-2x}$ and $g(x) = \log \left( \frac{2x+3}{3-2x} \right)$.Therefore, $g(x) = \log(f(x))$, which implies $g(1) = \log(f(1))$.

If $\int e^x \cos x \, dx = \frac{e^x}{2}(\cos x + \sin x)$ and $\int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx = \frac{f(x)}{24}[\cos (g(x)) + \sin (g(x))] + c$, then $g(1) =$

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