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∫ ( X + 1 ) √ X 2 − X + 1 D X - Mathematics

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प्रश्न

\[\int\left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
योग

उत्तर

\[\text{ Let I } = \int \left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
\[\text{ Also }, x + 1 = \lambda\frac{d}{dx}\left( x^2 - x + 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 2x - 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 2x - 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \left( 2\lambda \right)x + \mu - \lambda\]
\[\text{Equating the coefficient of like terms}\]
\[2\lambda = 1\]
\[ \Rightarrow \lambda = \frac{1}{2}\]
\[\text{ And }\]
\[\mu - \lambda = 1\]
\[ \Rightarrow \mu - \frac{1}{2} = 1\]
\[ \Rightarrow \mu = \frac{3}{2}\]
\[ \therefore I = \int\left[ \frac{1}{2}\left( 2x - 1 \right) + \frac{3}{2} \right] \sqrt{x^2 - x + 1}dx\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1}dx + \frac{3}{2}\int\sqrt{x^2 - x + 1}dx\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx}+ \frac{3}{2}\int \sqrt{x^2 - x + \frac{1}{4} - \frac{1}{4} + 1} \text{ dx}\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx} + \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \frac{3}{4}}\text{ dx}\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx}+ \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\text{ dx}\]
\[\text{ Let x}^2 - x + 1 = t\]
\[ \Rightarrow \left( 2x - 1 \right)dx = dt\]
\[ \therefore I = \frac{1}{2}\int\sqrt{t} \text{ dt }+ \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} dx\]
\[ = \frac{1}{2} \times \left( \frac{t^\frac{3}{2}}{\frac{3}{2}} \right) + \frac{3}{2}\left[ \frac{x - \frac{1}{2}}{2} \sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log } \left| x - \frac{1}{2} + \sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| \right] + C\]
\[ = \frac{1}{3} \left( x^2 - x + 1 \right)^\frac{3}{2} + \frac{3}{8}\left( 2x - 1 \right) \sqrt{x^2 - x + 1} + \frac{9}{16}\text{ log } \left| x - \frac{1}{2} + \sqrt{x^2 - x + 1} \right| + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.29 [पृष्ठ १५८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.29 | Q 1 | पृष्ठ १५८

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