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$\int_{1}^{e} \log x d x=$
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We have, $\int_{1}^{e} \log x d x$
$=\left[\log x \int 1 d x\right]_{1}^{e}-\int_{1}^{e}\left(\frac{d}{d x} \log x \int 1 d x\right) d x$
$=[\log x \cdot x]_{1}^{e}-\int_{1}^{e} \frac{1}{x} \cdot x d x$
$=[e \log e-1 \log (1)]-\int_{1}^{e} 1 d x$
$=e-[x]_{1}^{e}=e-[e-1]=1$
$=\left[\log x \int 1 d x\right]_{1}^{e}-\int_{1}^{e}\left(\frac{d}{d x} \log x \int 1 d x\right) d x$
$=[\log x \cdot x]_{1}^{e}-\int_{1}^{e} \frac{1}{x} \cdot x d x$
$=[e \log e-1 \log (1)]-\int_{1}^{e} 1 d x$
$=e-[x]_{1}^{e}=e-[e-1]=1$
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