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

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Question

(B) Let $I = \int 2^x (f^{\prime}(x) + f(x) \log 2) \, dx$. We can rewrite the integrand as: $I = \int (2^x f^{\prime}(x) + 2^x \log 2 \cdot f(x)) \,

Vedclass · Accepted Answer

$2^x f(x) + C$

Vedclass · Answer

(B) Let $I = \int 2^x (f^{\prime}(x) + f(x) \log 2) \, dx$. We can rewrite the integrand as: $I = \int (2^x f^{\prime}(x) + 2^x \log 2 \cdot f(x)) \, dx$. Recall the product rule for differentiation: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$. Let $u(x) = 2^x$ and $v(x) = f(x)$. Then $u^{\prime}(x) = \frac{d}{dx} (2^x) = 2^x \log 2$. Thus, the integrand is of the form $u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$, which is $\frac{d}{dx} [2^x f(x)]$. Therefore, $I = \int \frac{d}{dx} [2^x f(x)] \, dx = 2^x f(x) + C$.

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

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