Question

\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]

Answer 1

\[Let\ I = \int_0^1 \frac{1 - x}{1 + x} d\ x\ . Then, \]
\[I = \int_0^1 \left( \frac{1}{1 + x} - \frac{1 + x - 1}{1 + x} \right) d x\]
\[I = \int_0^1 \left( \frac{1}{1 + x} - 1 + \frac{x}{1 + x} \right) d x\]
\[ \Rightarrow I = \left[ \log \left( 1 + x \right) - x + \log \left( 1 + x \right) \right]_0^1 \]
\[ \Rightarrow I = \left( \log 2 - 1 + \log 2 \right) - \left( \log 1 - 0 + \log 1 \right)\]
\[ = 2 \log 2 - 1\]