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प्रश्न
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
योग
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उत्तर
\[\text{Let I} = \int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)}dx\]
\[\text{Putting}\ \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}}dx = dt\]
\[ \therefore I = \int\frac{1}{2 + 3t}dt\]
\[ = \frac{1}{3} \text{ln }\left| 2 + 3t \right| + C\]
\[ = \frac{1}{3} \text{ln }\left| 2 + 3 \sin^{- 1} x \right| + C \left[ \because t = \sin^{- 1} x \right]\]
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