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Question
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Solution
\[Let\ I = \int_1^3 \frac{\log x}{\left( 1 + x \right)^2} d\ x\ . Then, \]
\[I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 - \int_1^3 \frac{1}{x}\left( \frac{- 1}{x + 1} \right) d x\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \int_1^3 \frac{1}{x\left( x + 1 \right)} dx\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \int_1^3 \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \left[ \log x - \log \left( x + 1 \right) \right]_1^3 \]
\[ \Rightarrow I = \frac{- 1}{4} \log 3 + \log 3 - \log 4 + \log 2\]
\[ \Rightarrow I = \frac{3}{4} \log 3 - \log 2\]
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