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प्रश्न
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
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उत्तर
\[\text{ Let I } = \int\frac{\left( 3x + 1 \right) dx}{\sqrt{5 - 2x - x^2}}\]
\[\text{ Consider, }3x + 1 = A \frac{d}{dx} \left( 5 - 2x - x^2 \right) + B\]
\[ \Rightarrow 3x + 1 = A \left( - 2 - 2x \right) + B\]
\[ \Rightarrow 3x + 1 = \left( - 2A \right) x - 2A + B\]
\[\text{ Equating Coefficients of like terms }\]
\[ - 2A = 3\]
\[ \Rightarrow A = - \frac{3}{2}\]
\[\text{ And }\]
\[ - 2A + B = 1\]
\[ \Rightarrow - 2 \times - \frac{3}{2} + B = 1\]
\[ \Rightarrow B = 1 - 3\]
\[ \Rightarrow B = - 2\]
\[ \therefore I = \int\left[ \frac{- \frac{3}{2} \left( - 2 - 2x \right) - 2}{\sqrt{5 - 2x - x^2}} \right]dx\]
\[ = - \frac{3}{2}\int\frac{\left( - 2 - 2x \right) dx}{\sqrt{5 - 2x - x^2}} - 2\int\frac{dx}{\sqrt{5 - 2x - x^2}}\]
\[ = - \frac{3}{2}\int\frac{\left( - 2 - 2x \right) dx}{\sqrt{5 - 2x - x^2}} - 2\int\frac{dx}{\sqrt{5 - \left( x^2 + 2x \right)}}\]
\[ = - \frac{3}{2}\int\frac{\left( - 2 - 2x \right) dx}{\sqrt{5 - 2x - x^2}} - 2\int\frac{dx}{\sqrt{5 - \left( x^2 + 2x + 1 - 1 \right)}}\]
\[ = - \frac{3}{2}\int\frac{\left( - 2 - 2x \right) dx}{\sqrt{5 - 2x - x^2}} - 2\int\frac{dx}{\sqrt{6 - \left( x + 1 \right)^2}}\]
\[ = - \frac{3}{2}\int\frac{\left( - 2 - 2x \right) dx}{\sqrt{5 - 2x - x^2}} - 2\int\frac{dx}{\sqrt{\left( \sqrt{6} \right)^2 - \left( x + 1 \right)^2}}\]
\[\text{ let 5 - 2x - x^2 = t }\]
\[ \Rightarrow \left( - 2 - 2x \right) dx = dt\]
\[ \therefore I = - \frac{3}{2}\int\frac{dt}{\sqrt{t}} - 2\int\frac{dx}{\sqrt{\left( \sqrt{6} \right)^2 - \left( x + 1 \right)^2}}\]
\[ = - \frac{3}{2} \times 2\sqrt{t} - 2 \sin^{- 1} \left( \frac{x + 1}{\sqrt{6}} \right) + C\]
\[ = - 3\sqrt{5 - 2x - x^2} - 2 \sin^{- 1} \left( \frac{x + 1}{\sqrt{6}} \right) + C\]
