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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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Verified Answer
The correct answer is:
$e^{2}\left(1+\log _{e} 2\right)-e$
Let $I=\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x$
$\Rightarrow \quad I=\int_{1}^{2}\left(e^{x} \cdot \log _{e} x+e^{x}+\frac{e^{x}}{x}\right) d x$
$\Rightarrow \quad I=\int_{1}^{2} e^{x} \log _{e} x d x+\int_{1}^{2} e^{x} d x+\int_{1}^{2} \frac{e^{x}}{x} d x$
$-\int_{1}^{2} e^{x} \log _{e} x d x$
$\Rightarrow \quad I=\left(e^{2}-e^{1}\right)+\left(e^{2} \log _{e} 2-0\right)$
$=e^{2}\left(1+\log _{e} 2\right)-e$
$\Rightarrow \quad I=\int_{1}^{2}\left(e^{x} \cdot \log _{e} x+e^{x}+\frac{e^{x}}{x}\right) d x$
$\Rightarrow \quad I=\int_{1}^{2} e^{x} \log _{e} x d x+\int_{1}^{2} e^{x} d x+\int_{1}^{2} \frac{e^{x}}{x} d x$
$-\int_{1}^{2} e^{x} \log _{e} x d x$
$\Rightarrow \quad I=\left(e^{2}-e^{1}\right)+\left(e^{2} \log _{e} 2-0\right)$
$=e^{2}\left(1+\log _{e} 2\right)-e$
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