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प्रश्न
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
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उत्तर
\[\int_0^1 co t^{- 1} \left( 1 - x + x^2 \right) d x\]
\[ = \int_0^1 co t^{- 1} \left[ x\left( x - 1 \right) + 1 \right] d x\]
\[ = \int_0^1 co t^{- 1} \left[ \frac{\left( x\left( x - 1 \right) + 1 \right)}{x - \left( x - 1 \right)} \right] d x\]
\[ = \int_0^1 co t^{- 1} x - co t^{- 1} \left( x - 1 \right) dx\]
\[ = \left[ xco t^{- 1} x \right]_0^1 + \int_0^1 \frac{x}{1 + x^2}dx - \left[ \left( x - 1 \right)co t^{- 1} \left( x - 1 \right) \right]_0^1 - \int_0^1 \frac{\left( x - 1 \right)}{1 + \left( x - 1 \right)^2}dx\]
\[ = \left[ xco t^{- 1} x \right]_0^1 + \frac{1}{2} \left[ \log\left( 1 + x^2 \right) \right]_0^1 - \left[ \left( x - 1 \right)co t^{- 1} \left( x - 1 \right) \right]_0^1 - \frac{1}{2} \left[ \log\left( 1 + \left( 1 - x \right)^2 \right) \right]_0^1 \]
\[ = \frac{\pi}{4} - \frac{1}{2}\log2 + \frac{\pi}{4} - \frac{1}{2}\log2\]
\[ = \frac{\pi}{2} - \log2\]
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