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प्रश्न
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
योग
उत्तर
\[\text{ Let I} = \int\frac{1}{\sqrt{3 - 2x - x^2}}dx\]
\[ = \int\frac{1}{\sqrt{3 - \left( x^2 + 2x + 1 - 1 \right)}}dx\]
\[ = \int\frac{1}{\sqrt{4 - \left( x + 1 \right)^2}}dx\]
\[\text{ Putting} \left( x + 1 \right) = t\]
\[ \Rightarrow dx = dt\]
\[ \therefore I = \int\frac{dt}{\sqrt{2^2 - t^2}}\]
\[ = \sin^{- 1} \left( \frac{t}{2} \right) + C .................\left[ \because \int \frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C .....................\left[ \because t = \left( x + 1 \right) \right]\]
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