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Question

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

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Solution

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

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Definite Integrals

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Chapter 19: Definite Integrals - Exercise 20.1 [Page 16]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12

Chapter 19 Definite Integrals
Exercise 20.1 | Q 9 | Page 16

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