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Question
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
Solution
\[\int_0^1 \frac{1 - x}{1 + x} dx\]
\[ = \int_0^1 \frac{1 - x - 1 + 1}{1 + x} d x\]
\[ = \int_0^1 \frac{2 - \left( x + 1 \right)}{1 + x} d x\]
\[ = \int_0^1 \frac{2}{1 + x} - \int_0^1 \frac{1 + x}{1 + x}dx\]
\[ = \int_0^1 \frac{2}{1 + x} - \int_0^1 dx\]
\[ = 2 \left[ \log\left( 1 + x \right) \right]_0^1 - \left[ x \right]_0^1 \]
\[ = 2\log2 - 1\]
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