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

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प्रश्न

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

\[\text{ Let I } = \int\frac{\left( x + 1 \right) dx}{\sqrt{4 + 5x - x^2}}\]
\[\text{ Also,} x + 1 = A \frac{d}{dx} \left( 4 + 5x - x^2 \right) + B\]
\[x + 1 = A \left( 5 - 2x \right) + B\]
\[x + 1 = \left( - 2A \right) x + 5A + B\]
\[\text{Equating Coefficients of like terms}\]
\[ - 2A = 1\]
\[ \Rightarrow A = - \frac{1}{2}\]
\[\text{ And }\]
\[5A + B = 1\]
\[ \Rightarrow - \frac{5}{2} + B = 1\]
\[B = \frac{7}{2}\]
\[I = \int\frac{\left( x + 1 \right) dx}{\sqrt{4 + 5x - x^2}}\]
\[ = \int\left( \frac{- \frac{1}{2} \left( 5 - 2x \right) + \frac{7}{2}}{\sqrt{4 + 5x - x^2}} \right)dx\]
\[ = - \frac{1}{2}\int\frac{\left( 5 - 2x \right) dx}{\sqrt{4 + 5x - x^2}} + \frac{7}{2}\int\frac{dx}{\sqrt{4 - \left( x^2 - 5x \right)}}\]
\[ = - \frac{1}{2}\int\frac{\left( 5 - 2x \right) dx}{\sqrt{4 + 5x - x^2}} + \frac{7}{2}\int\frac{dx}{\sqrt{4 - \left[ x^2 - 5x + \left( \frac{5}{2} \right)^2 - \left( \frac{5}{2} \right)^2 \right]}}\]
\[ = - \frac{1}{2}\int\frac{\left( 5 - 2x \right) dx}{\sqrt{4 + 5x - x^2}} + \frac{7}{2}\int\frac{dx}{\sqrt{4 - \left( x - \frac{5}{2} \right)^2 + \frac{25}{4}}}\]
\[ = - \frac{1}{2}\int\left( \frac{5 - 2x}{\sqrt{4 + 5x - x^2}} \right)dx + \frac{7}{2}\int\frac{dx}{\sqrt{\frac{41}{4} - \left( x - \frac{5}{2} \right)^2}}\]
\[ = - \frac{1}{2}\int\left( \frac{5 - 2x}{\sqrt{4 + 5x - x^2}} \right)dx + \frac{7}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{41}}{2} \right)^2 - \left( x - \frac{5}{2} \right)^2}}\]
\[\text{ let } 4 + 5x - x^2 = t\]
\[ \Rightarrow \left( 5 - 2x \right) dx = dt\]
\[\text
{Then }, \]
\[I = - \frac{1}{2}\int\frac{dt}{\sqrt{t}} + \frac{7}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{41}}{2} \right)^2 - \left( x - \frac{5}{2} \right)^2}}\]
\[ = - \frac{1}{2} \times 2\sqrt{t} + \frac{7}{2} \times \sin^{- 1} \left( \frac{x - \frac{5}{2}}{\frac{\sqrt{41}}{2}} \right) + C\]
\[ = - \sqrt{t} + \frac{7}{2} \sin^{- 1} \left( \frac{2x - 5}{\sqrt{41}} \right) + C\]
\[ = - \sqrt{4 + 5x - x^2} + \frac{7}{2} \sin^{- 1} \left( \frac{2x - 5}{\sqrt{41}} \right) + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.21 [पृष्ठ ११०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.21 | Q 3 | पृष्ठ ११०

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