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