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1 ∫ 0 Tan − 1 ( 2 X 1 − X 2 ) D X - Mathematics

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प्रश्न

\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
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उत्तर

\[\int_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) d x\]
\[ = \int_0^1 2 \tan^{- 1} x\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - 2 \int_0^1 \frac{x}{1 + x^2}dx\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - \left[ \log\left( 1 + x^2 \right) \right]_0^1 \]
\[ = 2\frac{\pi}{4} - 0 - \log2 + 0\]
\[ = \frac{\pi}{2} - \log2\]

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Definite Integrals
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पाठ 20: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.2 | Q 17 | पृष्ठ ३९

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