The integral int_{1}^{2} e^{x} cdot x^{x}(1 + | vedclass

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(D) Let $I = \int_{1}^{2} e^{x} x^{x} (2 + \log_{e} x) dx$. We can rewrite the integrand as: $I = \int_{1}^{2} e^{x} [x^{x} + x^{x}(1 + \log_{e} x)] d

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$e(4e - 1)$

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(D) Let $I = \int_{1}^{2} e^{x} x^{x} (2 + \log_{e} x) dx$. We can rewrite the integrand as: $I = \int_{1}^{2} e^{x} [x^{x} + x^{x}(1 + \log_{e} x)] dx$. Let $f(x) = x^{x}$. Then $f'(x) = \frac{d}{dx}(e^{x \log_{e} x}) = e^{x \log_{e} x} \cdot \frac{d}{dx}(x \log_{e} x) = x^{x} (1 \cdot \log_{e} x + x \cdot \frac{1}{x}) = x^{x} (1 + \log_{e} x)$. Thus,the integral is of the form $\int_{1}^{2} e^{x} [f(x) + f'(x)] dx$. The result of this integral is $[e^{x} f(x)]_{1}^{2}$. Substituting $f(x) = x^{x}$: $I = [e^{x} x^{x}]_{1}^{2} = (e^{2} \cdot 2^{2}) - (e^{1} \cdot 1^{1}) = 4e^{2} - e = e(4e - 1)$.

The integral $\int_{1}^{2} e^{x} \cdot x^{x}(1 + \log_{e} x + 1) dx$ is equal to:

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