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∫ 1 √ 7 − 3 X − 2 X 2 D X - Mathematics

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प्रश्न

\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
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उत्तर

\[\int\frac{dx}{\sqrt{7 - 3x - 2 x^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \frac{3}{2}x - x^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \left( x^2 - \frac{3}{2}x \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x^2 + \frac{3}{2}x + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x + \frac{3}{4} \right)^2 + \frac{9}{16}}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} + \frac{9}{16} - \left( x + \frac{3}{4} \right)^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{56 + 9}{16} - \left( x + \frac{3}{4} \right)^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{65}}{4} \right)^2 - \left( x + \frac{3}{4} \right)^2}}\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left[ \frac{x + \frac{3}{4}}{\frac{\sqrt{65}}{4}} \right] + C\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left[ \frac{4x + 3}{\sqrt{65}} \right] + C\]

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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 6 | पृष्ठ ९३

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