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Question
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Solution
\[Let\ I = \int_1^2 \log\ x\ d\ x\ . Then, \]
\[I = \int_1^2 1 \log x\ d\ x\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 \frac{1}{x} x\ d\ x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 d x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \left[ x \right]_1^2 \]
\[ \Rightarrow I = 2 \log 2 - 2 + 1\]
\[ \Rightarrow I = 2 \log 2 - 1\]
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