int e^{2x + log x} dx = | vedclass

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(A) We are given the integral $I = \int e^{2x + \log x} dx$. Using the property of exponents $e^{a+b} = e^a \cdot e^b$ and $e^{\log x} = x$,we get: $I

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$\frac{1}{4}(2x - 1)e^{2x} + c$

Vedclass · Answer

(A) We are given the integral $I = \int e^{2x + \log x} dx$. Using the property of exponents $e^{a+b} = e^a \cdot e^b$ and $e^{\log x} = x$,we get: $I = \int e^{2x} \cdot e^{\log x} dx = \int x e^{2x} dx$. Now,apply integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = x$ and $v = e^{2x}$. Then $u' = 1$ and $\int v dx = \frac{e^{2x}}{2}$. $I = x \cdot \frac{e^{2x}}{2} - \int 1 \cdot \frac{e^{2x}}{2} dx$. $I = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + c$. $I = \frac{2x e^{2x} - e^{2x}}{4} + c = \frac{e^{2x}}{4}(2x - 1) + c$.

$\int e^{2x + \log x} dx = $

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