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प्रश्न
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
उत्तर
\[\int_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) d x\]
\[Let, x = \tan\theta,\text{ then }dx = se c^2 \theta d\theta\]
\[\text{When, }x \to 0 ; \theta \to 0\]
\[\text{And }x \to 1 ; \theta \to \frac{\pi}{4}\]
Therefore the integral becomes
\[ \int_0^\frac{\pi}{4} \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right) se c^2 \theta d\theta\]
\[ = \int_0^\frac{\pi}{4} \tan^{- 1} \left( \tan2\theta \right) se c^2 \theta d\theta\]
\[ = 2 \int_0^\frac{\pi}{4} \theta se c^2 \theta d\theta\]
\[ = 2 \left[ \theta \tan\theta \right]_0^\frac{\pi}{4} - 2 \int_0^\frac{\pi}{4} \tan\theta d\theta\]
\[ = 2 \left[ \theta \tan\theta \right]_0^\frac{\pi}{4} - 2 \left[ - \log\left( \cos\theta \right) \right]_0^\frac{\pi}{4} \]
\[\]
\[ = 2\left( \frac{\pi}{4} - 0 \right) + 2\left[ \log\frac{1}{\sqrt{2}} - 0 \right]\]
\[ = \frac{\pi}{2} - \log2\]
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