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1 ∫ 0 Log ( 1 + X ) D X

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प्रश्न

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

योग
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उत्तर

\[\int_0^1 \log\left( 1 + x \right) d x\]

\[ = \int_0^1 \log\left( 1 + x \right) \times 1 d x\]

\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \frac{x}{1 + x}dx\]

\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \left( 1 - \frac{1}{1 + x} \right)dx\]

\[ = \left[ x\log\left( 1 + x \right) \right]_0^1 - \left[ x - \log\left( 1 + x \right) \right]_0^1 \]

\[ = \log2 - 1 + \log2\]

\[ = 2\log2 - 1\]

\[ = \log4 - \log e\]

\[ = \log\frac{4}{e}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 22 | पृष्ठ १२१

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