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प्रश्न
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
पर्याय
1
0
−1
π/4
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उत्तर
0
\[Let\, I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} d x ................(1)\]
\[ = \int_0^1 \tan^{- 1} \frac{2\left( 1 - x \right) - 1}{1 + \left( 1 - x \right) - \left( 1 - x \right)^2} d x\]
\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{2 - x - 1 - x^2 + 2x} dx\]
\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{1 + x - x^2} dx\]
\[ = - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx .................(2)\]
\[\text{Adding (i) and (ii)}\]
\[2I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx\]
\[ = 0\]
\[Hence\, I = 0\]
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