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∫ 1 √ 16 − 6 X − X 2 D X - Mathematics

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प्रश्न

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
बेरीज
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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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पाठ 19: Indefinite Integrals - Exercise 19.17 [पृष्ठ ९३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.17 | Q 7 | पृष्ठ ९३

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