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∫ 1 √ 8 + 3 X − X 2 D X - Mathematics

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प्रश्न

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
योग

उत्तर

\[\int\frac{dx}{\sqrt{8 + 3x - x^2}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x - \frac{3}{2} \right)^2 + \frac{9}{4}}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{\left( \frac{\sqrt{41}}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2}}\]
\[ \Rightarrow \sin^{- 1} \left( \frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right) + C\]
\[ \Rightarrow \sin^{- 1} \left( \frac{2x - 3}{\sqrt{41}} \right) + C\]

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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 2 | पृष्ठ ९३

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