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प्रश्न
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
योग
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उत्तर
\[\text{ Let I } = \int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) \text{ dx }\]
\[ = \int3 \tan^{- 1} \left( x \right) \text{ dx }\]
\[ = 3\int\left[ \tan^{- 1} \left( x \right) \times 1 \right] \text{ dx }\]
\[ = 3 \left[ \tan^{- 1} x \times x - \int\frac{1}{1 + x^2} \times\text{ x dx } \right]\]
\[ = 3x \tan^{- 1} x - 3\int\frac{x}{1 + x^2} dx\]
\[\text{ let 1 }+ x^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[\text{ Then,} \]
\[I = 3x \tan^{- 1} x - \frac{3}{2}\int\frac{dt}{t}\]
\[ = 3x \tan^{- 1} x - \frac{3}{2} \text{ log } \left| t \right| + C\]
\[ = 3x \tan^{- 1} x - \frac{3}{2} \text{ log} \left| 1 + x^2 \right| + C\]
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