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Question
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Solution
\[\int_0^1 \log\left( 1 + x \right) d x\]
\[ = \int_0^1 \log\left( 1 + x \right) \times 1 d x\]
\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \frac{x}{1 + x}dx\]
\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \left( 1 - \frac{1}{1 + x} \right)dx\]
\[ = \left[ x\log\left( 1 + x \right) \right]_0^1 - \left[ x - \log\left( 1 + x \right) \right]_0^1 \]
\[ = \log2 - 1 + \log2\]
\[ = 2\log2 - 1\]
\[ = \log4 - \log e\]
\[ = \log\frac{4}{e}\]
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