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प्रश्न
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
योग
उत्तर
\[\text{We have}, \]
\[I = \int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[ = \int\frac{2 - \left( 1 - x^2 \right)}{\sqrt{1 - x^2}} \text{ dx }\]
\[ = 2\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }- \int\frac{1 - x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[ = 2\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }- \int\sqrt{1 - x^2} \text{ dx }\]
\[ = 2 \text{ sin}^{- 1} x - \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2} \sin^{- 1} x \right] + C\]
\[ = 2 \sin^{- 1} x - \frac{x}{2}\sqrt{1 - x^2} - \frac{1}{2} \sin^{- 1} x + C\]
\[ = \frac{3}{2} \text{ sin}^{- 1} x - \frac{x}{2}\sqrt{1 - x^2} + C\]
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