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∫ 1 √ 5 − 4 X − 2 X 2 D X - Mathematics

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प्रश्न

\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
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उत्तर

\[\int\frac{dx}{\sqrt{5 - 4x - 2 x^2}}\]
\[ = \int\frac{dx}{\sqrt{2\left[ \frac{5}{2} - 2x - x^2 \right]}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x + 1 - 1 \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x + 1 \right)^2 + 1}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \frac{\left( x + 1 \right)\sqrt{2}}{\sqrt{7}} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \sqrt{\frac{2}{7}}\left( x + 1 \right) \right) + C\]

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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 3 | पृष्ठ ९३

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