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प्रश्न

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
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उत्तर

\[\int\frac{dx}{\sqrt{7 - 6x - x^2}}\]
\[ = \int\frac{dx}{\sqrt{7 - \left( x^2 + 6x \right)}}\]
\[ = \int\frac{dx}{\sqrt{7 - \left[ x^2 + 6x + 3^2 - 3^2 \right]}}\]
\[ = \int\frac{dx}{\sqrt{7 + 9 - \left( x + 3 \right)^2}}\]
\[ = \int\frac{dx}{\sqrt{4^2 - \left( x + 3 \right)^2}}\]
\[ = \sin^{- 1} \left( \frac{x + 3}{4} \right) + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.17 [पृष्ठ ९३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.17 | Q 8 | पृष्ठ ९३

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