int 2^{x} [f^{prime}(x) + f(x) log 2] , dx is | vedclass

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Question

(C) Let $I = \int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$.We know that the derivative of the product of two functions is given by the product rule:

Vedclass · Accepted Answer

$2^{x} f(x) + C$

Vedclass · Answer

(C) Let $I = \int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$.We know that the derivative of the product of two functions is given by the product rule: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$.Consider the function $g(x) = 2^{x} f(x)$.Differentiating $g(x)$ with respect to $x$ using the product rule:$g^{\prime}(x) = \frac{d}{dx}(2^{x}) \cdot f(x) + 2^{x} \cdot \frac{d}{dx}(f(x))$$g^{\prime}(x) = 2^{x} \log 2 \cdot f(x) + 2^{x} f^{\prime}(x)$$g^{\prime}(x) = 2^{x} [f^{\prime}(x) + f(x) \log 2]$.Since the integrand is exactly the derivative of $g(x)$, we have:$I = \int g^{\prime}(x) \, dx = g(x) + C = 2^{x} f(x) + C$.

$\int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$ is equal to

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