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Question

\[\int\limits_0^1 \log\left( 1 + x \right) dx\]

Sum
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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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Definite Integrals
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Chapter 20: Definite Integrals - Revision Exercise [Page 121]

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RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Revision Exercise | Q 22 | Page 121

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