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∫ 6 X + 5 √ 6 + X − 2 X 2 Dx - Mathematics

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प्रश्न

\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]
बेरीज

उत्तर

\[\int\frac{\left( 6x + 5 \right) dx}{\sqrt{6 + x - 2 x^2}}\]
\[\text{ Let  6x + 5 = A}\frac{d}{dx}\left( 6 + x - 2 x^2 \right) + B\]
\[ \Rightarrow 6x + 5 = A \left( - 4x + 1 \right) + B\]
\[ \Rightarrow 6x + 5 = - 4A \text{ x }+ \left( A + B \right)\]
\[\text{Equating coefficients of like terms}\]
\[ - 4A = 6\]
\[ \Rightarrow A = - \frac{3}{2}\]
\[ \text{ and}\ A + B = 5\]
\[ \Rightarrow - \frac{3}{2} + B = 5\]
\[ \Rightarrow B = 5 + \frac{3}{2}\]
\[ \Rightarrow B = \frac{13}{2}\]
\[\text{ Then, 6x + 5 }= - \frac{3}{2} \left( - 4x + 1 \right) + \frac{13}{2}\]
\[ \therefore \int\frac{\left( 6x + 5 \right)}{\sqrt{6 + x - 2 x^2}}\text{ dx } = \int\left( \frac{\frac{- 3}{2}\left( - 4x + 1 \right) + \frac{13}{2}}{\sqrt{6 + x - 2 x^2}} \right)\text{ dx }\]
\[ = - \frac{3}{2}\int\frac{\left( - 4x + 1 \right)}{\sqrt{6 + x - 2 x^2}} \text{ dx }+ \frac{13}{2}\int\frac{1}{\sqrt{6 + x - 2 x^2}}\text{ dx }\]
\[\text{  Putting  6 + x - 2 x}^2 =\text{  t   in   the  Ist  integral}\]
\[ \Rightarrow \left( - 4x + 1 \right) \text{ dx } = dt\]
\[ \therefore \int\frac{\left( 6x + 5 \right)}{\sqrt{6 + x - 2 x^2}}\text{ dx }= - \frac{3}{2}\int\frac{1}{\sqrt{t}}dt + \frac{13}{2 \times \sqrt{2}}\int\frac{1}{\sqrt{3 + \frac{x}{2} - x^2}}\text{ dx }\]
\[ = - \frac{3}{2}\int t^{- \frac{1}{2}} \cdot dt + \frac{13}{2\sqrt{2}}\int\frac{1}{\sqrt{3 + \frac{x}{2} - x^2 - \left( \frac{1}{4} \right)^2 + \left( \frac{1}{4} \right)^2}}\text{ dx }\]
\[ = - \frac{3}{2}\int t^{- \frac{1}{2}} \cdot dt + \frac{13}{2\sqrt{2}}\int\frac{1}{\sqrt{3 + \frac{1}{16} - \left( x - \frac{1}{4} \right)^2}}\text{ dx }\]
\[ = - \frac{3}{2}\int t^{- \frac{1}{2}} \cdot dt + \frac{13}{2\sqrt{2}}\int\frac{1}{\sqrt{\left( \frac{7}{4} \right)^2 - \left( x - \frac{1}{4} \right)^2}}\text{ dx }\]
\[ = - 3 \left[ t^\frac{1}{2} \right] + \frac{13}{2\sqrt{2}} \times \sin^{- 1} \left( \frac{x - \frac{1}{4}}{\frac{7}{4}} \right) + C .............\left[ \because \int\frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = - 3 \sqrt{6 + x - 2 x^2} + \frac{13}{2\sqrt{2}} \times \sin^{- 1} \left( \frac{x - \frac{1}{4}}{\frac{7}{4}} \right) + C\]
\[ = - 3\sqrt{6 + x - 2 x^2} + \frac{13}{2\sqrt{2}} \sin^{- 1} \left( \frac{4x - 1}{7} \right) + C\]

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पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 75 | पृष्ठ २०४

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