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प्रश्न
\[\int x^2 \tan^{- 1} x\text{ dx }\]
योग
उत्तर
\[\text{ Let I } = \int {x^2}_{II} . \tan^{- 1}_1 \text{ x dx }\]
\[ = \tan^{- 1} x\int x^2 dx - \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int x^2 dx \right\}\text{ dx }\]
\[ = \tan^{- 1} x \times \frac{x^3}{3} - \int \left( \frac{1}{1 + x^2} \right) \times \frac{x^3}{3} \text{ dx }\]
\[ = \tan^{- 1} x\int x^2 dx - \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int x^2 dx \right\}\text{ dx }\]
\[ = \tan^{- 1} x \times \frac{x^3}{3} - \int \left( \frac{1}{1 + x^2} \right) \times \frac{x^3}{3} \text{ dx }\]
\[ = \tan^{- 1} x. \frac {x^3}{3} - \frac{1}{3}\int \frac{x^2 . x}{1 + x^2}dx\]
\[\text{ Let 1 }+ x^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[ \Rightarrow \text{ x dx }= \frac{dt}{2}\]
\[ \therefore I = \tan^{- 1} x . \frac{x^3}{3} - \frac{1}{6}\int \frac{\left( t - 1 \right)}{t} . dt\]
\[ = \tan^{- 1} x . \frac{x^3}{3} - \frac{1}{6}\int dt + \frac{1}{6}\int \frac{dt}{t}\]
\[ = \tan^{- 1} x . \frac{x^3}{3} - \frac{t}{6} + \frac{1}{6}\text{ log }\left| t \right| + C\]
\[ = \tan^{- 1} x . \frac{x^3}{3} - \frac{\left( 1 + x^2 \right)}{6} + \frac{1}{6}\text{ log }\left( 1 + x^2 \right) + C\]
\[ = \tan^{- 1} x . \frac{x^3}{3} - \frac{x^2}{6} + \frac{1}{6}\text{ log }\left( 1 + x^2 \right) - \frac{1}{6} + C\]
\[ = \tan^{- 1} x . \frac{x^3}{3} - \frac{x^2}{6} + \frac{1}{6}\text{ log }\left| 1 + x^2 \right| +\text{ C' where C' = C -} \frac{1}{6}\]
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