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प्रश्न
उत्तर
\[\text{ Let I } = \int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[ = 2\int 1_{II} . \tan^{- 1} x_I \text{ dx }\]
\[ = 2 \left[ \tan^{- 1} x\int1 \text{ dx }- \int\left\{ \frac{d}{dx}\left\{ \tan^{- 1} x \right\}\int1 \text{ dx }\right\}dx \right]\]
\[ = 2\left[ \tan^{- 1} x . x - \int\frac{1}{1 + x^2} \times \text{ x dx } \right]\]
\[ = 2 \tan^{- 1} x . x - \int \frac{2x}{1 + x^2} \text{ dx }\]
\[\text{ Putting 1 + x}^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[ \therefore I = 2x \tan^{- 1} x - \int \frac{dt}{t}\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| t \right| + C\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| 1 + x^2 \right| + C \left[ \because t = 1 + x^2 \right]\]
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