The value of the integral int_{1}^{2} e^{x}le | vedclass

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(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.Let $f(x) = \log_{e} x$. Then $f'(x) = \frac{1}{x}$.The integral can be rewritten as:

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$e^{2}\left(1+\log _{e} 2\right)-e$

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(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.Let $f(x) = \log_{e} x$. Then $f'(x) = \frac{1}{x}$.The integral can be rewritten as:$I = \int_{1}^{2} e^{x} \left( \log_{e} x + \frac{1}{x} + 1 \right) dx$$I = \int_{1}^{2} e^{x} \left( (\log_{e} x + \frac{1}{x}) + 1 \right) dx$This does not directly fit the form. Let us expand:$I = \int_{1}^{2} e^{x} \log_{e} x dx + \int_{1}^{2} e^{x} dx + \int_{1}^{2} \frac{e^{x}}{x} dx$Using integration by parts for $\int e^{x} \log_{e} x dx$:Let $u = \log_{e} x$, $dv = e^{x} dx$. Then $du = \frac{1}{x} dx$, $v = e^{x}$.$\int e^{x} \log_{e} x dx = e^{x} \log_{e} x - \int \frac{e^{x}}{x} dx$.Substituting this back into $I$:$I = [e^{x} \log_{e} x - \int \frac{e^{x}}{x} dx] + [e^{x}]_{1}^{2} + \int_{1}^{2} \frac{e^{x}}{x} dx$$I = [e^{x} \log_{e} x]_{1}^{2} + [e^{x}]_{1}^{2}$$I = (e^{2} \log_{e} 2 - e^{1} \log_{e} 1) + (e^{2} - e^{1})$Since $\log_{e} 1 = 0$:$I = e^{2} \log_{e} 2 + e^{2} - e$$I = e^{2}(1 + \log_{e} 2) - e$.

The value of the integral $\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x$ is

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