1 ∫ − 1 1 X 2 + 2 X + 5 D X - Mathematics

Question

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

Solution

\[Let\ I = \int_{- 1}^1 \frac{1}{x^2 + 2x + 5} d x . Then, \]
\[ I = \int_{- 1}^1 \frac{1}{\left( x^2 + 2x + 1 \right) + 4} d x\]
\[ \Rightarrow I = \int_{- 1}^1 \frac{1}{\left( x + 1 \right)^2 + 2^2} d x\]
\[ \Rightarrow I = \frac{1}{2} \left[ \tan^{- 1} \frac{\left( x + 1 \right)}{2} \right]_{- 1}^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \tan^{- 1} 1 - \tan^{- 1} 0 \right)\]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow I = \frac{\pi}{8}\]

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Definite Integrals

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Q 44 Q 43 Q 45

Chapter 20: Definite Integrals - Exercise 20.1 [Page 17]

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RD Sharma Mathematics [English] Class 12

Chapter 20 Definite Integrals
Exercise 20.1 | Q 44 | Page 17

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