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प्रश्न
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
बेरीज
उत्तर
\[\int\frac{dx}{\sqrt{16 - 6x - x^2}}\]
\[ = \int\frac{dx}{\sqrt{16 - \left( x^2 + 6x \right)}}\]
\[ = \int\frac{dx}{\sqrt{16 - \left( x^2 + 6x + 3^2 - 3^2 \right)}}\]
\[ = \int\frac{dx}{\sqrt{16 + 9 - \left( x + 3 \right)^2}}\]
\[ = \int\frac{dx}{\sqrt{5^2 - \left( x + 3 \right)^2}}\]
\[ = \sin^{- 1} \left( \frac{x + 3}{5} \right) + C\]
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